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Minchanka [31]
3 years ago
10

Please help me! Thank you! I only need help with a)

Mathematics
1 answer:
Rzqust [24]3 years ago
8 0

Since we know AD, CD and angle D, we can use the law of cosines, which states that

AC^2 = AD^2+CD^2 - 2AD\cdot CD\cos(D)

Plugging the values, we have

AC^2 = 17^2+13^2 - 2\cdot 17\cdot 13\cos(42)

Which leads to

AC^2 = 458 - 442\cos(42) \approx 129.530

Taking the square root, we have

AC \approx \sqrt{129.530} \approx 11.3811

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A math test is worth 100 points and has 38 problems. Each problem is worth either 5 points or 2 points. How many problems of eac
Nat2105 [25]

Answer:

<em><u>5</u></em><em><u> </u></em><em><u>points</u></em><em><u> </u></em><em><u>question</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>8</u></em>

<em><u>and</u></em><em><u> </u></em><em><u>2</u></em><em><u> </u></em><em><u>po</u></em><em><u>ints</u></em><em><u> questions</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u> </u></em><em><u>!</u></em><em><u>!</u></em>

Step-by-step explanation:

Let the number of 5 points questions be x and number of the 2 points questions be y !!

Given :- The paper has 38 problems

i.e,

number of 5 points Q + number of 2point Q = 38

=> x + y = 38

=> x = 38 - y .... ( i )

And , The test worth 100 pts :-

There must be x number of Questions which carry 5 points and there must y number of questions which carries 2 points ( as assume above )

so, 5x + 2y = 100 .... ( ii )

Putting value of the x in ( ii )

5 ( 38 - y ) + 2y = 100

190 - 5y + 2y = 100

- 3y = - 90

y = 30

Putting value of y in ( i )

x = 38 - y

x = 38 - 30

x = 8

So, the Number of 5points Question is

x = 8

and , the number of 2 points questions is y = 30 !!

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Answer: 1 2/3 more

Step-by-step explanation:

6 0
3 years ago
Choose the equation that has solutions (5,7) and (8,13).
Alekssandra [29.7K]
<h2>Answer:y=2x-3</h2>

Step-by-step explanation:

(x_{1},y_{1}) is called solution of a equation if (x_{1},y_{1}) satisfies the equation.

Consider the equation y=2x-3,

For point (5,7),

LHS=y=7

RHS=2x-3=2(5)-3=10-3=7

So,RHS=LHS

The point (5,7) satisfies the equation.

For point (8,13),

LHS=y=13

RHS=2x-3=2(8)-3=16-3=13

So,RHS=LHS

The point (8,13) satisfies the equation.

So,both the points satisfy the equation y=2x-3

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