Answer quick PLS<br><br> i will give brainliest

suppose that O has coordinates (0,0),and the points P and Q have coordinates that are whole numbers between 0 and 2, inclusive.

pishuonlain [190]

Answer:

Step-by-step explanation:

suppose that O has coordinates (0,0),and the points P and Q have coordinates that are whole numbers between 0 and 2, inclusive. One example of a triangle with O,P, and Q as vertices is shown below . how many such triangle are right triangle ?

4 0

3 years ago

HI I really need help with this its in the picture

Hitman42 [59]

<h3>Answer: 24a^4 + 36a^3 + 12a^2</h3>

Explanation:

The blue rectangle has area of 4a^2*6a^2 = 24a^4

The red rectangle has area 4a^2*9a = 36a^3

The green rectangle has area 4a^2*3 = 12a^2

The total area is 24a^4 + 36a^3 + 12a^2

7 0

2 years ago

Jose equation is -4(x+2) = 8x+16 says to jill “I can solve for the variable in this equation in at least 4 different ways

erastovalidia [21]

Answer:

that is step 1 and step 2 and step3

step 4

Let's solve your equation step-by-step.

−4(x+2)=8x+16

Step 1: Simplify both sides of the equation.

−4(x+2)=8x+16

(−4)(x)+(−4)(2)=8x+16(Distribute)

−4x+−8=8x+16

−4x−8=8x+16

Step 2: Subtract 8x from both sides.

−4x−8−8x=8x+16−8x

−12x−8=16

Step 3: Add 8 to both sides.

−12x−8+8=16+8

−12x=24

Step 4: Divide both sides by -12.

−12x

−12

=

24

−12

x=−2

Answer:

x=−2

6 0

3 years ago

Simplify ( 8∙4∙2 8∙7 )^2 × ( 8 0 7−3 )^3 × 7 −9 .

stich3 [128]

$\bf ~\hspace{7em}\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
~\hspace{4.5em}
a^n\implies \cfrac{1}{a^{-n}}
~\hspace{4.5em}
\cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\left( \cfrac{8\cdot 4\cdot 2}{8\cdot 7} \right)^2\times \left( \cfrac{8^0}{7^{-3}} \right)^3\times 7^{-9}\implies \left( \cfrac{8\cdot 8}{8\cdot 7} \right)^2\times \left( \cfrac{1\cdot 7^3}{1} \right)^3\times \cfrac{1}{7^9}$

$\bf \left( \cfrac{8}{8}\cdot \cfrac{8}{7} \right)^2\times (7^3)^3\times \cfrac{1}{7^9}\implies \left( \cfrac{8}{7} \right)^2\times 7^{3\cdot 3}\times \cfrac{1}{7^9}\implies \cfrac{8^2}{7^2}\times \cfrac{7^9}{7^9}
\\\\\\
\cfrac{8^2}{7^2}\implies \cfrac{64}{49}$