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2 years ago

Cindy runs 13 blocks to the park and 13 blocks back home every day. If she does this 4 days, how many blocks will she run?

Mathematics

1 answer:

o-na [289]2 years ago

8 0

Answer:

I believe the answer is 104

Step-by-step explanation:

13+13=26 thats 1 day so 26×4=104

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Use the fundamental identities and appropriate algebraic operations to simplify the following expression. (18 +tan x) (18-tan x)

andrezito [222]

Answer:

a) $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)=325$

b) The lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

Step-by-step explanation:

a) To simplify the expression $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)$ you must:

Apply Difference of Two Squares Formula: $\left(a+b\right)\left(a-b\right)=a^2-b^2$

$a=18,\:b=\tan \left(x\right)$

$\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)=18^2-\tan ^2\left(x\right)=324-\tan ^2\left(x\right)$

$324-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Apply the Pythagorean Identity $1+\tan ^2\left(x\right)=\sec ^2\left(x\right)$

From the Pythagorean Identity, we know that $1=-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Therefore,

$324[-\tan ^2\left(x\right)+\sec ^2\left(x\right))]\\324[+1]\\325$

b) According with the below graph, the lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

3 0

3 years ago

Use logarithmic differentiation to find dy/dx

liq [111]

Answer:

dy/dx = (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]

Step-by-step explanation:

y = (x^2 - 3)^sinx

ln y = ln (x^2 - 3)^sinx

ln y = sin x * ln (x^2 - 3)

1/y * dy/dx = sin x * {1 / (x^2 - 3)} * 2x + ln(x^2 - 3) * cos x

1/y dy/dx = 2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)

dy/dx = [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)] * y

dy/dx = (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]

7 0

3 years ago

What is a correct congruence statement for the triangles shown? Enter your answer in the box. △QDJ≅△ A triangle Q D J. The b

almond37 [142]

Answer: The correct congruence statement is $\triangle QDJ\cong \triangle MCW$ .

Explanation:

It is given that A triangle Q D J. The base D J is horizontal and side Q D is vertical. Another triangle M C W is made. The base C W and side M C are neither horizontal nor vertical. Triangle M C W is to the right of triangle Q D J.

The sides Q D and M C are labeled with a single tick mark. The sides D J and C W are labeled with a double tick mark. The sides Q J and M W are labeled with a triple tick mark.

Draw two triangles according to the given information.

From the figure it is noticed that

$QD\cong MC$