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Maksim231197 [3]
2 years ago
7

Find X to the nearest tenth.A) 9.3B) 9.4C) 9.5

Mathematics
2 answers:
hodyreva [135]2 years ago
7 0

Answer:

Choice B is correct answer.

Step-by-step explanation:

From question statement, we observe that

Perpendicular = x

Hypotenuse = 10

∅  = 70°

We have to find the value of x.

Since we know that sin∅= perpendicular/hypotenuse

Putting the values in above formula,we get

sin70°= x/10

Multiplying above equation by 10,we get

10sin70° = (x/10)10

10sin70° = x

x = 10×.9396

x = 9.396≈9.4

Hence, the value of x is 9.4

antoniya [11.8K]2 years ago
4 0

Answer:

Second choice

Step-by-step explanation:

The unknown leg x in right triangle is opposite to the angle of the measure 70°.

Use the definition of the sine of the angle:

\sin 70^{\circ}=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{x}{10}.

Hence,

x=10\sin 70^{\circ}\approx 9.4.

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7 0
2 years ago
A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee
IrinaVladis [17]

Answer:

720 ways

Step-by-step explanation:

Generally, combination is expressed as;

                                  ^{n} C_{r} = \frac{n!}{r!(n-r)!}

The question consists of 9 multiple-choice questions and examinee must answer 7 of the multiple-choice questions.

                                   ⇒ ⁹C₇ =\frac{9!}{7!(9-7)!}

                                        =\frac{9!}{7!(2)!}

                                        = 36

The question consists of 6 open-ended problems and examinee must answer 3 of the open-ended problems.

                                    ⇒ ⁶C₃ =\frac{6!}{3!(6-3)!}

                                         =\frac{6!}{3!(3)!}

                                         = 20

Combining the two combinations to determine the number of ways the questions and problems be chosen if an examinee must answer 7 of the multiple-choice questions and 3 of the open-ended problem.

                                        ⁹C₇ × ⁶C₃

                                       = 36 × 20

                                       = 720 ways  

3 0
3 years ago
use the function g=f+4 to find the value of g when f=1use the function u=10c to find the value of u when c=5how about this one u
NeTakaya

Given function g=f+4

To find the value of g at f=1

substitute f=1 in the given equation

\begin{gathered} g=f+4 \\ g=1+4 \\ g=5 \end{gathered}

So, at f=1, the value of g will be 5.

(b).

At u=10c

to find the value of u at c=5

substitute the value c=5 in the given equation

\begin{gathered} u=10c \\ u=10\times5 \\ u=50 \end{gathered}

The value of u at c=5 is 50

(c).

Given function u=n-5

to find the value of u at n=7

substitute the value of n=7 in the given equation

\begin{gathered} u=n-5 \\ u=7-5 \\ u=2 \end{gathered}

So the value of u at n=7 is 2

(d).

Given function h=g+13

to find the value of h at g=1

substitute the value g=1 in the given equation

\begin{gathered} h=g+13 \\ h=1+13 \\ h=14 \end{gathered}

The value of h at g=1 is 14

(e).

Given function w=14f

to find the value of w at f=4

substitute the value f=4 in the given equation

\begin{gathered} w=14f \\ w=14\times4 \\ w=56 \end{gathered}

The value of w at f=4 is 56

4 0
1 year ago
A builder needs to add diagonal braces to a wall. The wall is 16 feet wide by 12 feet high. What is the length of each brace?
bagirrra123 [75]
12 16 20 those are it i think<span />
3 0
2 years ago
Read 2 more answers
Marissa can buy 3 oranges for $1.35 4 oranges for $.80 or 5 oranges for $2.25 marissa can buy 10 oranges for a price that fits t
agasfer [191]

3 oranges= $1.35, so one orange cost $0.45

4 oranges= $1.80, one orange cost $0.45

5 oranges= $2.25, one orange cost $0.45

If one orange cost $0.45, then ten cost $4.50 ($0.45*10)


If you were to buy ten oranges are $4.25, this would make each orange $0.425 ($0.43 rounded).


This means that if you were to buy ten oranges, it would be cheaper to buy ten for $4.25, instead of buying two lots of 5 for $2.25 each.



Read more on Brainly.com - brainly.com/question/302454#readmore

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