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3 years ago

Find the perimeter of a shape with the coordinates (-3, 3), (-1, 6), (2, 4), and (-1, 3)

Mathematics

1 answer:

Lerok [7]3 years ago

3 0

I believe it is (5,16)
.
honestly im not sure if this is right just my best educated guess sorry.
.
Zane

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hoa [83]

$\begin{gathered} F(x)\text{ is }increa\sin g\text{ on (}0,\infty\text{), beacuse the derivative is positive on that } \\ \text{Interval} \end{gathered}$

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1 year ago

Can anyone please help me with this

alekssr [168]

Answer:

Step-by-step explanation:

The correct explanation is A

$\frac{x^{m} }{a^{n} }$ = $a^{(m-n)}$

Thus

$\frac{9.2}{4}$ × $\frac{10^{8} }{10^{-3} }$

= 2.3 × $10^{(8-(-3))}$

= 2.3 × $10^{11}$

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3 years ago

I need help with this question on my hw

PtichkaEL [24]

Answer:

A) 630 feet squared

Step-by-step explanation:

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3 years ago

Please Help me, what the answer

Rus_ich [418]

It’s either 1. Or 2. I’m really not sure which one tho so sorry! There’s not enough information given but if I had to guesstimate I would say 1.

6 0

3 years ago

The coordinates of A, B, and C in the diagram are A(p,4), B(6,1), and C(9,q). Which equation correctly relates p and q?

siniylev [52]

Answer:

$q+p=7$

Step-by-step explanation:

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

In this problem line AB and line BC are perpendicular

$m_A_B*m_B_C=-1$

step 1

Find the slope AB

we have

A(p,4), B(6,1)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{1-4}{6-p}$

$m_A_B=-\frac{3}{6-p}$

step 2

Find the slope BC

we have

B(6,1), and C(9,q)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{q-1}{9-6}$

$m_B_C=\frac{q-1}{3}$

step 3

Find the equation that relates p and q

we know that

$m_A_B*m_B_C=-1$

we have

$m_A_B=-\frac{3}{6-p}$

$m_B_C=\frac{q-1}{3}$

substitute

$(-\frac{3}{6-p})(\frac{q-1}{3})=-1$

$(\frac{q-1}{6-p})=1$

$q-1=6-p\\q+p=7$

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3 years ago