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3 years ago

Geometry Homework... Have been absent and trying to catch up. Any help is appreciated :)

Mathematics

1 answer:

strojnjashka [21]3 years ago

5 0

B because you need to know if those angles are congruent

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1. Find p(-3) if p(x) = 4x3 - 5x2 + 7x - 10

SVETLANKA909090 [29]

Answer:

1) -184

2) 13 $(m^{4}) ^{2}$ -3( $m^{4}$ )+4

Step-by-step explanation:

For problem number 1, Let's start by substituting all the x's for -3

So the equation will look like

p(-3)= $4(-3)^{3}$ - $5(-3)^{2}$ +7(-3)-10

Now, lets simplify by solving the exponents

First we'll do $-3^{3}$ which equals -27

Now we'll do $-3^{2}$ which equals 9

So now our equation looks like

p(-3)=4(-27)-5(9)+7(-3)-10

Now, we are going to multiply all these numbers

First 4(-27) which equals -108

Now -5(9) which equals -45

And finally 7(-3) which equals -21

Now our equation looks like

p(-3)= -108-45-21-10

Now we can solve

-108-45-21-10=-184

8 0

3 years ago

How do you find the probability? give me details please and answer my question step by step.

mart [117]

Answer:

to find theoretical probability, devide the favorable outcomes by the total possible outcomes

to find experimental probability, for example on a coin, devide how many times it landed on heads by how many times you flipped it

i hope this helps :)

8 0

3 years ago

ΔABC has three sides with these lengths: AB = 9, BC = 40, and CA = 41. What is the value of cos C?

kotykmax [81]

Answer:

$cos(C) =\frac{40}{41}\\\\cos(C)=0.976$

Step-by-step explanation:

To find the $cos (C)$ we must use the cosine theorem.

The cosine theorem says that:

$c^2 = a^2 +b^2 -2abcos(C)$

In this case:

$c=AB = 9$

$b=CA = 41$

$a=BC = 40$

$9^2 = 40^2 +41^2 -2(40)*(41)cos(C)$

$9^2- 40^2- 41^2 =-2(40)*(41)cos(C)$

$cos(C) = -\frac{9^2- 40^2- 41^2}{2(40)*(41)}$

$cos(C) =\frac{40}{41}\\\\cos(C)=0.976$

7 0

3 years ago

When a person is breathing normally the amount of air in their lawns varies sinusoidally. When full Karen’s lungs hold 2.8 L of

makkiz [27]

Answer:

$A(t) = 2.2\sin \frac{(t - 2)\pi }{6} + 0.6$

Step-by-step explanation:

Let the function of quantity in the lung of air be A(t)

So $A(t) \alpha \sin (\frac{t - \alpha }{k} )$

so, A(t) = Amax sin t + b

A(t) = 2.8t⇒ max

A(t) = 0.6t ⇒ min

max value of A(t) occur when sin(t) = 1

and min value of A(t) = 0

So b = 0.6

and A(max) = 2.2

$A(t) = 2.2\sin \frac{(t)}{k} + 0.6$

at t = 2 sec volume of a is 0.6

So function reduce to

$A(t) = 2.2\sin \frac{(t - 2)}{k} + 0.6$

and t = 5 max value of volume is represent

so,

$\sin \frac{t - \alpha }{k} = 1$

$\frac{t - 2}{k} = \frac{\pi }{2}$ when t = 5

$\frac{6}{\pi } = k$

so the equation becomes

$A(t) = 2.2\sin \frac{(t - 2)\pi }{6} + 0.6$

7 0

3 years ago

2x + 8 = 2(x + 4) help pls

Fynjy0 [20]

2x +8=2x+8
2x-2x=8-8
0x=0
So the answer is 0

7 0

3 years ago

Read 2 more answers