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2 years ago

12

Solve X/3+5=9 pls give me answer

1 answer:

dalvyx [7]2 years ago

5 0

The answer is x= 12.

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Write an expression that is equivalent to three over four(5z + 16).

klio [65]

Your answer is B. Using the distributive property, 3/4 • 5z equals 15/4z, which can't be simplified. Then 3/4 of 16 is 12. If you divide 16 by 4 (the denominator) and then multiply by 3 (the numerator) that yields your answer.

8 0

3 years ago

Find the slope of a line perpendicular to 2x-y-16

Alekssandra [29.7K]

put the equation 2x - y = 16 in the form of y = mx + c

2x - y = 16

2x = 16 + y

y = 2x - 16

the slope of this line is 2. the slope of a line perpendicular to it would be the negative reciprocal of 2. in other words, it would multiply with 2 to give -1.

you can form this equation with that info

2x = -1

x = -1/2

OR

you can flip and change the sign (numerator) of 2/1

2/1

= -1/2

6 0

3 years ago

Which of the following functions is graphed below?

Irina18 [472]

Answer:

fat bit

fat old ugly

Step-by-step explanation:

5 0

2 years ago

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

3 years ago

The sequence of transformations that can be performed on quadrilateral ABCD to show that it is congruent to quadrilateral GHIj i

Andreas93 [3]

Answer:

1).

A. Reflection across the y axis.

2).

A. Clockwise rotation about point B and a translation 20 units down.

3 0

3 years ago

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