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2 years ago

If X= -6 and y=4 what is x² + y²

Mathematics

2 answers:

Rainbow [258]2 years ago

6 0

Answer:

Step-by-step explanation:

(-6)x(-6)=36

(4)x(4)=16

36+16=52

Jobisdone [24]2 years ago

4 0

Answer:

x = 6 x 6

y = 4 x 4

Step-by-step explanation:

This is because...

x^2 means "x cubed", meaning multiply it by itself. Same for y.

Now the math...

6 x 6 = 36

4 x 4 = 16

Math 2

Without rounding, we find the true sum of 36 + 16 = 52.

Therefore,

x^2 + y^2 = 52

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Solve by factoring p² + 16p +64 = 0 i like highkey don’t know how to do this

bonufazy [111]

Answer:

-8

Step-by-step explanation:

When you factor, you just find two numbers that add to be the first number (16) and multiply to be the second number (64). 8 and 8 both do this. So when we factor, we also add the variable into each equation because it also has been divided into two equations, and we get: (p + 8)(p +8). Notice that the equations are "+8" and not "-8" because two +8s add to give us a +16 and multiply to give us a +64. After that, we just solve the two equations to find the solutions. Subtract 8 from each equation (we do the opposite operation because 8 is positive) to get -8. Because the answer for each equation is the same our answer is -8.

7 0

3 years ago

Read 2 more answers

ANSWER MY QUESTION PLEASE

mina [271]

When you plug the equation into a graphing calculator the x-intercept is -5
Or you can set the equation equal to 0 and solve for x.

6 0

3 years ago

Find the Range and Domain of the Binary relations

sukhopar [10]

Answer:

A = {3,5,6}

B = {1,3}

A x B = {(3,1),(3,3),(5,1),(5,3),(6,1),(6,3)}

Domain of A x B = (The x figures in ascending order)

=> {3,5,6} (Figures repeating more than once must be written only 1 time)

Range of A x B = (The y figures in ascending order)

=> {1,3}

B x A = {(1,3),(3,3),(1,5),(3,5),(1,6),(3,6)}

Domain of B x A = {1,3}

Range of A x B = {3,5,6}

8 0

3 years ago

Find the volume and area for the objects shown and answer Question

klio [65]

Step-by-step explanation:

You must write formulas regarding the volume and surface area of the given solids.

$\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=\dfrac{4}{3}\pi r^3\\SA=4\pi r^2$

$\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\$

$\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}$

$\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}$

$SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}$