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3 years ago

Name the missing coordinate(s) of the triangle

Mathematics

1 answer:

rewona [7]3 years ago

7 0

Answer:

R(7a, 0 )

Step-by-step explanation:

R is on the vertical line RQ so will have the same x- coordinate as Q

R is on the horizontal line OR so will have the same y- coordinate as O

Thus coordinates of R = (7a, 0 )

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Let f(x)=−12(x+2)2+5 . What is the average rate of change for the quadratic function from x=−3 to x = 1?

aev [14]

Answer:

Average rate of change = -6

Step-by-step explanation:

The average rate of change over the interval (a,b) is given by;

[ f(b) - f(a)] / (b-a)........................where interval is (a,b)

(a,b) interval =(-3,1)

Where;

f(x)= -12 (x+2)² +5

$f(a)=f(-3)=-12(-3+2)^2+5\\f(a)=f(-3)=-12(-1)^2+5\\f(a)=f(-3)=-12(1)+5\\=-12+5=-7\\\\\\f(b)=f(1)=-12(1+2)+5\\f(b)=f(1)=-12(3)+5\\=-36+5=-31\\\\f(b)-f(a)=-31--7=-24\\\\b-a=1--3=4\\\\f(b)-f(a)/b-a=\frac{-24}{4} =-6$

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2 years ago

Please help I’ll give brainliest

Hatshy [7]

Help with what tho you need help with what I don’t see it

4 0

3 years ago

Verify sin(180degrees-0)=sin0

egoroff_w [7]

Answer:

See Explanation

Step-by-step explanation:

$sin(180-0)=sin(180)\\sin(180)=0\\sin(0)=0\\0=0$

5 0

3 years ago

Solve as long multiplication / Show work.(So I can understand the steps) $655.84 x 0.062 =

Gennadij [26K]

Answer:

Step-by-step explanation:

1. 655.84

x 0.0062

131168

+ 393504

000000

0000000

0004066208

there are 5 numbers after the decimal point,we count from the back and place our point = 40.66208

8 0

3 years ago

Drag the tiles to the boxes to form correct pairs. Match the pairs of equivalent expressions.

Rashid [163]

Answer:

The following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Step-by-step explanation:

Lets solve all the expressions to match the results.

$5\left(2t+1\right)+\left(-7t+28\right)$

Solving the expression

$5\left(2t+1\right)+\left(-7t+28\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$5\left(2t+1\right)-7t+28$

$10t+5-7t+28$

$3t+33$

Therefore, $5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$

$3\left(3t-4\right)-\left(2t+10\right)$

Solving the expression

$3\left(3t-4\right)-\left(2t+10\right)$

$9t-12-\left(2t+10\right)$

$9t-12-2t-10$

$7t-22$

Therefore, $3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

Solving the expression

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)$

$4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)$

$4t-\frac{8}{5}-3+\frac{4t}{3}$

$-3-\frac{8}{5}:\quad -\frac{23}{5}$ and $\frac{4t}{3}+4t:\quad \frac{16t}{3}$

So,

$4t-\frac{8}{5}-3+\frac{4t}{3}$ will become $\frac{16t}{3}-\frac{23}{5}$

Therefore, $\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

Solving the expression

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$-\frac{9}{2}t+3+\frac{7}{4}t+33$

$\mathrm{Group\:like\:terms}$

$\frac{9}{2}t+\frac{7}{4}t+3+33$

$\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t$

$-\frac{11}{4}t+3+33$

$-\frac{11}{4}t+36$

Therefore, $\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Thus, the following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Keywords: algebraic expression

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5 0

3 years ago

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