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2 years ago

Find A, B, and C 7th Grade Math Please help me answer this

Mathematics

1 answer:

xeze [42]2 years ago

3 0

Answer:

A=124 B=60 and C=60

Hope this helps.

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What is one times one times four times twelve divided by three divided by two times nine minus twelve divided by twelve times on

xenn [34]

The answer is 617,283,945.

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3 years ago

What is the -2 squared -2

oksano4ka [1.4K]

Answer: the answer is negative one over four

-1/4

6 0

3 years ago

A card is randomly drawn from an ordinary deck of playing cards. What is the probability that the card is a king or a diamond?

Feliz [49]

Answer:

There are 52 possible outcomes when you draw one card from a standard deck. Hence, 52 is the denominator of your probability fraction.

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2 years ago

I Need Some Help!!!!!!!!!

ehidna [41]

Answer:

median = 50

difference in the quartiles = 45

Step-by-step explanation:

The median amount is the line in the middle of the box

The median is 50

The difference between the upper quartile and the lower quartile is the difference between the right end of the box and the left end of the box

65-20 = 45

4 0

3 years ago

Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len

Rama09 [41]

The length of a curve C parameterized by a vector function r(t) = x(t) i + y(t) j over an interval a ≤ t ≤ b is

$\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt$

In this case, we have

x(t) = exp(t ) + exp(-t ) ==> dx/dt = exp(t ) - exp(-t )

y(t) = 5 - 2t ==> dy/dt = -2

and [a, b] = [0, 2]. The length of the curve is then

$\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt$

$=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt$

$=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt$

$=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt$

$=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}$

5 0

2 years ago