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Sav [38]
4 years ago
8

Which of the following is a valid step in the solution?

Mathematics
2 answers:
Lena [83]4 years ago
8 0
C is the correct answer
kirill115 [55]4 years ago
6 0

Answer:

pretty sure the answer is the last one

Step-by-step explanation:

you put them to the 4th power to get rid of the root sign

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A vessel sails 41 miles S 35° E. How far south has it sailed?
const2013 [10]

Option A: The vessel has sailed 34 miles South

Explanation:

Given that the vessel sails 41 miles S 35° E.

We need to determine the how far south the vessel has sailed.

Let us use the Pythagorean theorem to find the distance of the vessel has sailed.

Thus, we have,

cos \ \theta=\frac{adj}{hyp}

Substituting \theta=35^{\circ} , adj=x and hyp=41

Hence, we get,

cos \ 35=\frac{x}{41}

Simplifying, we have,

0.82=\frac{x}{41}

Multiplying both sides by 41, we have,

33.62=x

Rounding off to the nearest integer, we have,

34=x

Thus, the vessel has sailed 34 miles South.

Therefore, Option B is the correct answer.

6 0
3 years ago
Help pls asap i’ll mark as brainliest if correct :)
xeze [42]

Answer:

100 chocolate bars

Step-by-step explanation:

30 ÷ 3 = 10 so you need to times 10 by 10?

hope that helps!

3 0
3 years ago
Read 2 more answers
Find x in the diagram
sukhopar [10]

Answer: where’s the diagram

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The area of a rectangular parking lot is represented by A = 6x^2 − 19x − 7 If x represents 15 m, what are the length and width o
Mekhanik [1.2K]

Answer:

The length and width of the parking lot are \frac{46}{3} meters and \frac{23}{2} meters, respectively.

Step-by-step explanation:

The surface formula (A) for the rectangular parking lot is represented by:

A = w\cdot l

Where:

w - Width of the rectangle, measured in meters.

l - Length of the rectangle, measured in meters.

Since, surface formula is a second-order polynomial, in which each binomial is associated with width and length. If A = 6\cdot x^{2}-19\cdot x -7, the factorized form is:

A = \left(x-\frac{7}{2}\,m \right)\cdot \left(x+\frac{1}{3}\,m \right)

Now, let consider that w = \left(x-\frac{7}{2}\,m \right) and l = \left(x+\frac{1}{3}\,m \right), if x = 15\,m, the length and width of the parking lot are, respectively:

w =\left(15\,m-\frac{7}{2}\,m \right)

w = \frac{23}{2}\,m

l =\left(15\,m+\frac{1}{3}\,m \right)

l = \frac{46}{3}\,m

The length and width of the parking lot are \frac{46}{3} meters and \frac{23}{2} meters, respectively.

5 0
3 years ago
Forth grade math: What happens to the area of the triangle when one of the dimensions is doubled and halved?
Alja [10]

\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=base\\ h=height\\[-0.5em] \hrulefill\\ b=\stackrel{doubled}{2b} \end{cases}\implies A=\cfrac{1}{2}(2b)(h)\implies \stackrel{\textit{twice as the original area}}{A=bh} \\\\[-0.35em] ~\dotfill

\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=base\\ h=height\\[-0.5em] \hrulefill\\ h=\stackrel{halved}{\frac{h}{2}} \end{cases}\implies A=\cfrac{1}{2}b\left( \cfrac{h}{2} \right)\implies \stackrel{\textit{half of the original area}}{A=\cfrac{1}{2}bh\left( \cfrac{1}{2} \right)}

7 0
4 years ago
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