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2 years ago

Which of the following numbers would be in the solution set to this inequality? Mark all that apply. -2x + 16 ≤ 10

1 answer:

4 0

-2x + 16 < = 10
-2x < = 10 - 16
-2x < = - 6
x > = -6/-2
x > = 3............so any number that is 3 or above is in the solution set

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Solve for r.<br> 686 = 14r

Naily [24]

Answer:

r = 49

Step-by-step explanation:

686 = 14r

686/14= 49

r = 49

8 0

2 years ago

Read 2 more answers

1. A square ABCD has the vertices A(n, n), B(n, -n), C(-n, -n), and D(-n, n). Which vertex is in Quadrant III?

andrew-mc [135]

The four quadrants and the sign of coordinates is given as follows:

Quadrant I : (n, n ) both x and y positive.

Quadrant II : ( -n,n) = x negative and y positive.

Quadrant III : ( -n,-n) = both x and y negative.

Quadrant IV: (n, -n) = x positive and y negative.

So Based on the above concept , we can say that vertex C (-n,-n) has both x and y negative and so it lies in quadrant III.

Answer is Vertex C (-n,-n)

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3 years ago

Read 2 more answers

Anyone know the answer to this algebra problem?

ikadub [295]

Answer: $\bold{a=1\qquad b=\dfrac{1}{16}\qquad c=\dfrac{1}{64}\qquad d=1\qquad e=\dfrac{4}{9}\qquad f=\dfrac{16}{81}}$

<u>Step-by-step explanation:</u>

$\begin{array}{c|l}\underline{\quad x\quad}&\underline{\quad 4^{-x}\qquad \qquad}\\-1&4^{-(-1)}=4^1=4\\\\0&4^{-(0)}=4^0=1\\\\2&4^{-(2)}=\dfrac{1}{4^2}=\dfrac{1}{16}\\\\4&4^{-(4)}=\dfrac{1}{4^4}=\dfrac{1}{64}\end{array}$

$\begin{array}{c|l}\underline{\quad \bigg x \quad}&\underline{\quad \bigg(\dfrac{2}{3}\bigg)^x\qquad \qquad}\\\\-1&\bigg(\dfrac{2}{3}\bigg)^{-1}=\dfrac{3}{2}\\\\0&\bigg(\dfrac{2}{3}\bigg)^{0}=1\\\\2&\bigg(\dfrac{2}{3}\bigg)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}\\\\4&\bigg(\dfrac{2}{3}\bigg)^{4}=\dfrac{2^4}{3^4}=\dfrac{16}{81}\end{array}$

3 0

2 years ago

How do I solve for x

Alchen [17]

Greetings!

To find the length of any side of a right triangle, you can use the Pythagorean Thereom. It states that the squares of two sides are equal to the square of the hypotenuse:
$a^2+b^2=c^2$

Input the information from the diagram into the formula:
$(x)^2+(x+7)^2=(13)^2$

Expand each term:
$(x)^2+(x+7)^2=(13)^2$

$x^2+((x+7)(x+7))=169$

$x^2+(x(x+7)+7(x+7))=169$

$x^2+x^2+7x+7x+49=169$

Combine like terms:
$2x^2+14x+49=169$

Add -169 to both sides:
$(2x^2+14x+49)+(-169)=(169)+(-169)$

$2x^2+14x-120=0$

Factor out the Common Term (2):
$2(x^2+7x-60)=0$

Factor the Complex Trinomial:
$2(x^2-5x+12x-60)=0$

$2(x(x-5)+12(x-5))=0$

$2(x-5)(x+12)=0$

Set Factors to equal 0:
$x-5=0$

$x=5$

or

$x+12=0$

$x=-12$

However, since we are solving for the side length, the only possible answer is 5 (a shape can't have a side with a negative length.)

The Solution Is:
$\boxed{x=5}$

I hope this helped!
-Benjamin

6 0

2 years ago

Two sides of an isosceles triangle are unknown. We can say both are X ft

Y_Kistochka [10]

Answer:

c.12

Step-by-step explanation:

you do 33-9 which gets you 24. You then divide that by 2 so you get 12.

3 0

2 years ago