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3 years ago

Need help pleasethank you

Mathematics

1 answer:

ad-work [718]3 years ago

8 0

To divide like bases just subtract the exponents
answer is 2^(13-3) or 2^10 or 1024

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Help with this geometry question.

KATRIN_1 [288]

Plug in 4m in the space

6 0

3 years ago

((MULTIPLE CHOICE ))

lesya692 [45]

Answer:

Neither.

Step-by-step explanation:

you are correct about the pattern, so it's not adding the same number and not multiplying by the same number. So it's neither.

6 0

3 years ago

Read 2 more answers

What is the approximate area of a sector given O= 56 degrees with a diameter of 12m?

jeyben [28]

Area of sector is 17.584 meters

<em><u>Solution:</u></em>

Given that we have to find the approximate area of a sector given O= 56 degrees with a diameter of 12m

Diameter = 12 m

Radius = Diameter / 2 = 6 m

An angle of 56 degrees is the fraction $\frac{56}{360}$ of the whole rotation

A sector of a circle with a sector angle of 56 degrees is therefore also the fraction $\frac{56}{360}$ of the circle

The area of the sector will therefore also be $\frac{56}{360}$ of the area

$\text{ sector area } = \frac{56}{360} \times \pi r^2\\\\\text{ sector area } = \frac{56}{360} \times 3.14 \times 6^2\\\\\text{ sector area } = 17.584$

Thus area of sector is 17.584 meters

3 0

3 years ago

What is the value of s in the equation 4(2s − 1) = 7s + 12? (4 points)

notka56 [123]

Answer:

B) 16

8s-4=7s+12

s-4=12

s=16

3 0

3 years ago

Read 2 more answers

A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters

In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

$\Rightarrow y=\sqrt{x^2+100}$

Differentiating with respect to t

$\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}$

Since the car driving towards the intersection at 13 m/s.

so, $\frac{dx}{dt}=-13$

$\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)$

Now

$\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)$

$=\frac{24\times (-13)}{\sqrt{676}}$

$=\frac{24\times (-13)}{26}$

= -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0

3 years ago