Rezolvati, in multimea numerelor intregi, inecuatiile
Exercitiul 8 va rog!

there are two pieces, both divided in 10 equal segments.

so each piece is 10/10, or namely 1 whole.

two pieces, 10/10 each, is 20/10 for the whole thing, but there are only 11 shaded, so if we take 20 to be the 100%, what is 11 off of it in percentage?

[tex]\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 20&100\\ 11&x \end{array}\implies \cfrac{20}{11}=\cfrac{100}{x}\implies 20x=1100\\\\\\ x=\cfrac{1100}{20}\implies x=55[/tex]

Answer:

$14

Step-by-step explanation:

Wages earned by daniel in one hour = $8

Number of working hours spend by daniel in each week = 35

Wages earned by daniel in each week without pay increase = [tex]35\times 8 = 280[/tex]

Increase in pay per week = [tex]\frac{5}{100} \times 280= 14[/tex]

Therefore extra wages earned by Daniel will be $14.

$14

Step-by-step explanation:

Wages earned by daniel in one hour = $8

Number of working hours spend by daniel in each week = 35

Wages earned by daniel in each week without pay increase =

Increase in pay per week =

Therefore extra wages earned by Daniel will be $14.

Answer:

• discriminant: 73

• # of real solutions: 2

Step-by-step explanation:

Comparing the equation ...

2x^2 -9x +1 = 0

to the generic form ...

ax^2 +bx +c = 0

we find the coefficient values to be ...

a = 2; b = -9; c = 1

That makes the value of the discriminant, (b^2 -4ac), be ...

(-9)^2 -4(2)(1) = 81 -8 = 73

Since the discriminant is positive, the number of real solutions is 2.

Answer:

1,-3,-5

Step-by-step explanation:

Given:

f(x)=x^3+7x^2+7x-15

Finding all the possible rational zeros of f(x)

p= ±1,±3,±5,±15 (factors of coefficient of last term)

q=±1(factors of coefficient of leading term)

p/q=±1,±3,±5,±15

Now finding the rational zeros using rational root theorem

f(p/q)

f(1)=1+7+7-15

   =0

f(-1)= -1 +7-7-15

     = -16

f(3)=27+7(9)+21-15

    =96

f(-3)= (-3)^3+7(-3)^2+7(-3)-15

     = 0

f(5)=5^3+7(5)^2+7(5)-15

    =320

f(-5)=(-5)^3+7(-5)^2+7(-5)-15

      =0    

f(15)=(15)^3+7(15)^2+7(15)-15

     =5040

f(-15)=(-15)^3+7(-15)^2+7(-15)-15

      =-1920

Hence the rational roots are 1,-3,-5 !

Answer:

Actual rational zeros of f(x) are x=-5, x=-3, and x=1.

Step-by-step explanation:

Given function is [tex]f(x)=x^3+7x^2+7x-15[/tex].

constant term = p = -15

coefficient of leading term q= 1

then possible rational roots of given function f(x) are the divisors of [tex]\pm \frac{p}{q}=\pm 1, \quad \pm 3, \quad \pm 5, \quad \pm 15[/tex]

Now we can plug those possible roots into given function to see which one of them gives output 0.

test for x=1

[tex]f(x)=x^3+7x^2+7x-15[/tex]

[tex]f(1)=1^3+7(1)^2+7(1)-15[/tex]

[tex]f(1)=1+7+7-15[/tex]

[tex]f(1)=0[/tex]

Hence x=1 is the actual rational zero.

Similarly testing other roots, we get final answer as:

Actual rational zeros of f(x) are x=-5, x=-3, and x=1.

Answer:

C) 110.

Step-by-step explanation:

To find it you must multiply in number by 5.

22*5=110.

To prove it you must divide it into 5, since 20 is the fifth part of 110.

110/5=22.

Answer:

Step-by-step explanation:

[tex]\bold{Method\ 1}\\\\\begin{array}{ccc}22&-&20\%\\x&-&100\%\end{array}\qquad\text{cross multiply}\\\\20x=(22)(100)\\20x=2200\qquad\text{divide both sides by 20}\\x=110[/tex]

[tex]\bold{Method\ 2}\\\\\begin{array}{cccc}22&-&20\%&\text{multiply both sides by 5}\\5\cdot22&-&5\cdot20\%\\110&-&100\%\end{array}[/tex]

[tex]\bold{Method\ 3}\\\\p\%=\dfrac{p}{100}\to 20\%=\dfrac{20}{100}=0.2\\\\n-number\\\\20\%\ of\ n\ is\ 22\to0.2x=22\qquad\text{divide both sides by 0.2}\\\\x=\dfrac{22}{0.2}\\\\x=\dfrac{220}{2}\\\\x=110[/tex]

Answer:

The ordered pair that satisfies this problem is (1, -6); x = 1 when y = -6.

Step-by-step explanation:

Please, rewrite your question as follows:

"Suppose that y varies directly with x and y = 12 when x = -2. What is x when

y = -6?"  The " = " sign must be included.

The pertinent proportional relationship is y = kx, where k is the constant of proportionality.

We must find k here.  Let y = 12 and x = -2.  Then 12 = k(-2), or k = -6.

Then the relationship is y = -6x.

Now let y = -6 and find x:      -6 = -6x, or x = 1.

The ordered pair that satisfies this problem is (1, -6)

M = 3

Is that what you were looking for?

Hope this helps. please add brainlist

Answer:

Step-by-step explanation:

(-)(-) = (+)

(+)(+) = (+)

(-)(+) = (+)(-) = (-)

--------------------------------------

16 - (-53) = 16 + 53 = 69

Answer:

The entry has an area of __44___ square feet

The total area of the tree house is __345__ square feet

Step-by-step explanation:

Let's first calculate the area of the trapezoid entrance:

The area of a trapezoid is given by:

A =((b1 + b2) * h) / 2

Where b1 and b2 are the bases or parallel sides.  So, we have:

A = ((6 + 16) * 4) / 2 = (22 * 4) / 2 = 44 sq ft

Now, let's look at the other areas... which are all rectangles, so easy to calculate (b * h).

Playroom: 16 x 14 = 224 sq ft

Side Deck: 3 x 14 = 42 sq ft

Back Porch: 6 x6 = 35 sq ft

So, in the total area of the tree house is:

TA = 44 + 224 + 42 + 35 = 345 sq ft

Answer:

The entry terrace has an area of

48

square feet. The total area of the tree house including the entrance, porch, and side deck is

350

square feet.

Step-by-step explanation:

I just took a test with this question and this was the right answer

~Please mark me as brainliest :)

Answer:

[tex]x=55[/tex]

Step-by-step explanation:

The given equation is [tex]\sqrt{x-6}+3=10[/tex]

Group similar terms to get:

[tex]\sqrt{x-6}=10-3[/tex]

[tex]\sqrt{x-6}=7[/tex]

Square both sides to obtain:

[tex]x-6=7^2[/tex]

[tex]x-6=49[/tex]

Add 6 to both sides

[tex]x=49+6[/tex]

Simplify the right hand side to obtain;

[tex]x=55[/tex]

Answer:

Step-by-step explanation:

[tex]\left\{\begin{array}{ccc}y=5x+3&(1)\\y=4x+6&(2)\end{array}\right\\\\\text{Substitute (1) to (2):}\\\\5x+3=4x+6\qquad\text{subtract 3 from both sides}\\5x=4x+3\qquad\text{subtract 4x from both sides}\\x=3\\\\\text{Put the value of x to (1):}\\\\y=5(3)+3\\y=15+3\\y=18[/tex]

Answer:

300cm

Step-by-step explanation:

First you get the length, width, and height. The length is 15, the width is 12, and the height is 10.

15*12=180

Since their is 2 you multiply by 2

180*2=360

12*10=120

Since their is 2 you multiply by 2

120*2=240

Then you divide by 2 since it is a triangle but before that you add the products up

360+240=600

600/2=300

Answer:

Option A is correct that It is a FALSE Statement.

Step-by-step explanation:

Given:

Total number of student = 1

Number of student in Physical Science = 0.3

Number of student in Chemistry = 0.35

Number of student in Biology = 0.35

Number of student who are Sophomores = 0.55

To find: False Statement among given ones.

Percentage of the student in Physical Science = [tex]\frac{0.3}{1}\times100[/tex] = 30%

Percentage of the student in Chemistry = [tex]\frac{0.35}{1}\times100[/tex] = 35%

Percentage of the student in Biology = [tex]\frac{0.35}{1}\times100[/tex] = 35%

Percentage of the student who are Sophomores = [tex]\frac{0.55}{1}\times100[/tex] = 55%

Therefore, Option A is correct that It is a FALSE Statement.

Answer:

A is the answer

Step-by-step explanation:

This ones a bit tricky but the answer is A

Answer:

[tex]x^{2}+y^{2} +4x-2y+1=0[/tex]

Step-by-step explanation:

we know that

The general form of the equation of a circle is

[tex]x^{2} +y^{2}+ Dx + Ey + F=0[/tex]

where D, E, F are constants

step 1

Find the radius of the circle

Remember that the distance from the center to any point on the circle is equal to the radius

so

the formula to calculate the distance between two points is equal to

[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]

[tex](-2,1)\\(-4,1)[/tex]  

substitute the values

[tex]r=\sqrt{(1-1)^{2}+(-4+2)^{2}}[/tex]

[tex]r=\sqrt{(0)^{2}+(-2)^{2}}[/tex]

[tex]r=2\ units[/tex]

step 2

Find the equation of the circle in standard form

[tex](x-h)^{2} +(y-k)^{2}=r^{2}[/tex]

In this problem we have

center ( -2,1)

radius r=2 units

substitute

[tex](x+2)^{2} +(y-1)^{2}=2^{2}[/tex]

[tex](x+2)^{2} +(y-1)^{2}=4[/tex]

Step 3

Convert to general form

[tex](x+2)^{2} +(y-1)^{2}=4\\ \\x^{2}+4x+4+y^{2}-2y+1=4\\ \\x^{2}+y^{2} +4x-2y+1=0[/tex]

Answer:

- [tex]\frac{2}{729}[/tex]

Step-by-step explanation:

The n th term of a geometric sequence is

[tex]a_{n}[/tex] = a₁ [tex](r)^{n-1}[/tex]

Given a₁ = 6 and r = - [tex]\frac{1}{3}[/tex], then

[tex]a_{8}[/tex] = 6 × [tex](-1/3)^{7}[/tex] = 6 × - [tex]\frac{1}{3^{7} }[/tex] = 6 × - [tex]\frac{1}{2187}[/tex] = - [tex]\frac{2}{729}[/tex]

Answer:

B.

Step-by-step explanation:

The line gets smaller left to right so that means the line is negative. Since the line is negative x must be negative. The line is also touching the y-axis at 6. We are going to use the formula y=mx+b. So our equation is going to be y=6-x.

Answer: The answer to 3.25÷0.5 is 6.5 :)

Step-by-step explanation:

Answer:

[tex]553\ acres[/tex]

Step-by-step explanation:

Let

z------> the crop yield

x----> the number of bushels harvested

y----> the number of acres used for the harvest

[tex]z=\frac{x}{y}[/tex]

we have

[tex]x=94,010\ bushels[/tex]

[tex]z=170\ bushels/acre[/tex]

substitute and solve for y

[tex]170=\frac{94,010}{y}[/tex]

[tex]y=\frac{94,010}{170}=553\ acres[/tex]

Answer:

553

Step-by-step explanation:

The standard form of the equation of the parabola with a focus at (5, 0) and a directrix at x = -5 is (x - 5)^2 = 20y.

The standard form of the equation of a parabola with a focus at (h, k) and a directrix at x = d is given by (x - h)^2 = 4p(y - k), where p is the distance between the vertex and the focus or directrix.

In this case, the focus is at (5, 0) and the directrix is at x = -5. Since the directrix is a vertical line, the parabola opens to the right. The distance between the vertex and the focus or directrix is given by p = |5 - (-5)|/2 = 5 units.

Substituting the values into the standard form equation gives (x - 5)^2 = 4(5)(y - 0), which simplifies to (x - 5)^2 = 20(y - 0). Therefore, the standard form of the equation of the parabola is (x - 5)^2 = 20y.

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If the cone are similar, the volume ratio of Cone A to Cone B will be 1:25.

The right circular cone is one in which the line from the cone's peak to the center of the circle's base is perpendicular to the base's surface.

Assume that the radius of the right circular cone under consideration is 'r' units, and the height 'h' units. Then the volume is then expressed as;

[tex]\rm V = \dfrac{1}{3} \pi r^3 h \: \rm unit^3[/tex]

If the cone are similar, the volume ratio of Cone A to Cone B;

[tex]\rm \frac{V_A}{V_B}=(\frac{R_A}{R_B})^3 } \\\\\ \frac{V_A}{V_B}=( \frac{5}{25} )^3 \\\\\ \frac{V_A}{V_B}=\frac{1}{125}[/tex]

If the cone are similar, the volume ratio of Cone A to Cone B will be 1:25.

Learn more about cone volume here:

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A would be your best option. Choice A properly displays the equation, 5x+6=4x+(-3), using words. Let me know if you want an explanation of why the other options are invalid. :)

Answer:

A is correct.

Step-by-step explanation:

Which is the correct way to model the equation 5x+6=4x+(-3) using algebra tiles?  

5 positive x-tiles and 6 positive unit tiles on the left side; 4 positive x-tiles and 3 negative unit tiles on the right side.

5 positive x-tiles = 5x

6 positive unit tiles = 6

4 positive x-tiles = 4x

3 negative unit tiles = -3

-----------------------------------------------------------------------

B - There are 5 x tiles but its given 6 x tiles.

C  - There is 6 positive unit tile and its given 6 negative,

D - There is 3 negative unit tile but here its given positive.

Answer:

[tex]\lim_{x \to 0} \frac{x^2}{sinx}=0[/tex]

Step-by-step explanation:

[tex]f(-0.03)=-0.0300\\f(-0.02)=-0.0200\\f(-0.01)=-0.0100\\f(0.01)=0.0100\\f(0.02)=0.0200\\f(0.03)=0.0300[/tex]

As the Left side of the function is approaching 0 as it gets closer to x=0 and the Right side is also approaching 0 as it gets closer to x=0, the limit would be 0.

Answer:

Step-by-step explanation:

-3 is the number u will get. Hope this helps!!

To represent the expression -8 + 5 on a number line, start at -8 and move 5 units to the right, ending at -3.

To represent the expression -8 + 5 on a number line, we start at -8 and move 5 units to the right. Since we are adding a positive number, the arrow on the number line will point to the right.

So the number line that represents the expression -8 + 5 would start at -8 and have an arrow pointing to the right, ending at -3.

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Answer: 32

Step-by-step explanation:

Answer:

Perimeter of triangle JKL: [tex]2 \sqrt{17} + 2\sqrt{34}[/tex].

Area of triangle JKL: 17.

Step-by-step explanation:

None of the three sides of triangle JKL is parallel to either the x-axis or the y-axis. Apply the Pythagorean Theorem to find the length of each side.

[tex]\rm JK = \sqrt{(3 - (-5))^{2} + (4- 6)^{2}} = \sqrt{8^{2} + (-2)^{2}} = \sqrt{68} = 2\sqrt{17}[/tex].

[tex]\rm JL = \sqrt{(-2 - (-5))^{2} + (1- 6)^{2}} = \sqrt{3^{2} + (-5)^{2}} = \sqrt{34}[/tex].

[tex]\rm KL = \sqrt{(-2 - 3)^{2} + (1-4)^{2}} = \sqrt{(-5)^{2} + (-3)^{2}} = \sqrt{34}[/tex].

The perimeter of triangle JKL will be:

[tex]\rm JK + JL + KL = 2\sqrt{17} + \sqrt{34} + \sqrt{34} = 2 \sqrt{17} + 2\sqrt{34}[/tex].

In case you realized that [tex]\rm JK : JL : KL = \sqrt{2} : 1 : 1[/tex], which makes JKL an isosceles right triangle:

Area of a right triangle:

[tex]\begin{aligned}\displaystyle \rm Area &= \frac{1}{2} \times \text{First Leg} \times \text{Second Leg}\\ &=\frac{1}{2} \times \sqrt{34}\times\sqrt{34}\\&= 17\end{aligned}[/tex].

Alternatively, apply the Law of Cosines to find the cosine of any of the three internal angles. This method works even if the triangle does not contain a right angle.

Taking the cosine of angle K as an example:

[tex]\displaystyle\begin{aligned}\rm \cos{K}&=\frac{(\text{First Adjacent Side})^{2} + (\text{Second Adjacent Side})^{2}-(\text{Opposite Side})^{2}}{2\times (\text{First Adjacent Side})\times(\text{Second Adjacent Side})}\\&\rm =\frac{(JK)^{2} + (JL)^{2} -(KL)^{2}}{2\times JK \times JL}\\&=\frac{(2\sqrt{17})^{2}+(\sqrt{34})^{2}-(\sqrt{34})^{2}}{2\times\sqrt{34} \times(2\sqrt{17})}\\ &=\frac{2^{2}\times 17}{2\times \sqrt{2}\times\sqrt{17}\times 2\times \sqrt{17}}\\&=\frac{1}{\sqrt{2}}\end{aligned}[/tex].

Apply the Pythagorean Theorem to find the sine of angle K:

[tex]\displaystyle \rm \sin{K} = \sqrt{1 - (\cos{K})^{2}} = \sqrt{1 - \left(\frac{1}{\sqrt{2}}\right)^{2} } = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}[/tex].

The height of JKL on the side JK will be:

[tex]\displaystyle \rm KL \cdot \sin{K} = \sqrt{34} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{68}}{2} = \frac{2\sqrt{17}}{2} = \sqrt{17}[/tex].

What will be the area of JKL given its height [tex]\sqrt{17}[/tex] on a base of length [tex]2\sqrt{17}[/tex]?

[tex]\displaystyle \rm Area = \frac{1}{2} \times Base\times Height = \frac{1}{2}\times (2\sqrt{17})\times \sqrt{17} = 17[/tex].

Step-by-step explanation:

The center of the circle is moved 3 units to the left and 4 units down.  So a = -3 and b = -4.

Among the given options, only ((-196)^(1/4)) is equivalent to a real number, as it involves the fourth root of a negative number, which results in a real value.

The correct answer is option B.

To determine which expression is equivalent to a real number, we need to consider the roots involved and whether they result in a real value.

Let's analyze each option:

A. (-1503)^(1/6): The expression involves the sixth root of a negative number. Odd roots of negative numbers are real, but even roots are not. Therefore, this expression is not guaranteed to be real.

B. (-196)^(1/4): The expression involves the fourth root of a negative number. Similar to option A, the fourth root of a negative number is real, so this expression is valid.

C. (-144)^(1/2): This expression involves the square root of a negative number. Square roots of negative numbers are not real; hence, this expression is not equivalent to a real number.

D. (-1024)^(1/2): The expression involves the square root of a negative number as well. Like option C, this expression is not equivalent to a real number.

In summary, option B ((-196)^(1/4)) is equivalent to a real number because it involves the fourth root of a negative number, and odd roots of negative numbers are real.

Answer:

c 8700

Step-by-step explanation:

Answer:

The answer is (C. 8700

Step-by-step explanation:

I took the test

The three methods are graphing, substitution and elimination.

In my opinion they are all easy but if I had to choose one it would be graphing because when I did homework assignments for this I always took the longest graphing than just solving the systems using substitution or elimination