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3 years ago

12

HELP RIGHT AWAY ASAP PLEASE!!!

1 answer:

Mumz [18]3 years ago

7 0

Answer:

when I worked it out I actually got -1466...

Step-by-step explanation:

Using order of operations you would multiply the 2 by 742, then subtract from left to right. 2 x 742 =1484 and 30 minus that is -1454. Subtracting 12 would give you -1466. I know that's not in the answer choices but that's what I got...

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Given the following diagram, enter the required information.

Verizon [17]

∠СXA = ∠BXA + ∠СXB = 30°20' + 40°35' = 70°55'

6 0

3 years ago

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

5 0

3 years ago

ANSWER ASAP GIVING BRAILIEST AND STUFF!

MrRissso [65]

The answer multiplied if it doubled the its times 2 tripled it’s timeS 3 quadruple then times 4

8 0

2 years ago

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What two variables are used most often in 2-variable equations?

MAXImum [283]

I believe it is x and y

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3 years ago

Hi! ❤️ , im looking for some help here. ill give brainliest if able to.

olchik [2.2K]

Hello there! The solution of your question has been given below.

<h3>Solution:</h3>

${6}^{ - 2} \\ = ( \frac{1}{6} ) ^{2} \\ = \frac{ {1}^{2} }{ {6}^{2} } \\ = \frac{1}{36}$

<h3>Answer:</h3>

Option B.

$\frac{1}{36}$

5 0

2 years ago