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GaryK [48]
2 years ago
13

The graph of an equation drawn through which two points would best represent the relationship between the number of miles and th

e cost of the trip? Cost and Distance (10,5) and (250, 120) (40, 10) and (250, 145) 160 150- 140- 130 120 - 110- 100- 90 80 (15, 25) and (240, 150) . 70 Cost (in dollars) 60 (60, 25) and (240, 150) 50 40 30 20 10 0 0 20 40 60 80 100 120 140 160 180 200 220 240 260 Distance (In miles) Done-​
Mathematics
1 answer:
andreyandreev [35.5K]2 years ago
5 0

Answer:

250

Step-by-step explanation:

Plz ve me brainlest

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If -2 is added to a number and the sum is doubled, the result is -15 less than the number.
VMariaS [17]

The required number is 19 given that -2 is added to a number, the sum is doubled and the result is -15 less than the number. This can be obtained by assuming the number as x, converting the given conditions to algebraic expression, forming algebraic equation and solving for x.

<h3>Find the required number:</h3>

Here in the question it is given that,

  • -2 is added to a number
  • The sum is doubled
  • The result is -15 less than the number

We have to find the required number.

Let the required number be x.

⇒ -2 is added to the number ⇒ -2 + x

⇒ the sum is doubled ⇒ 2(-2 + x)

⇒ the result is -15 less than the number ⇒ 2(-2 + x) = x -(-15)

2(-2 + x) = x + 15

⇒ - 4 + 2x = x + 15

⇒ x = 19

Hence the required number is 19 given that -2 is added to a number, the sum is doubled and the result is -15 less than the number.

Learn more about algebraic expression and equation here:

brainly.com/question/953809

#SPJ9

6 0
2 years ago
Explain how modeling partial products can be used to find the products of greater numbers ?
liq [111]

You can break large numbers into a sum of a multiple(s) of 10 and the last digit of the number. For example, you can break 26 as 20+6, or 157 as 100+50+7.

Then, using the distributive property, you can turn the original multiplication into a sum of easier multiplications. For example, suppose we want to multiply 26 and 37. This is quite challenging to do in your mind, but you can break the numbers as we said above:

26\times 37=(20+6)(30+7) = 20\times 30+20\times 7+30\times 6+6\times 7

All these multiplications are rather easy, because they either involve multiples of 10 of single-digit numbers:

20\times 30+20\times 7+30\times 6+6\times 7 = 600+140+180+42=962

3 0
3 years ago
If A and B are two angles in standard position in Quadrant I, find cos( A +B ) for the given function values. sin A = 8/17 and c
horsena [70]

Answer:

Part 1) cos(A + B) = \frac{140}{221}

Part 2) cos(A - B) = \frac{153}{185}

Part 3) cos(A - B) = \frac{84}{85}

Part 4) cos(A + B) = -\frac{36}{85}

Part 5) cos(A - B) = \frac{63}{65}

Part 6) cos(A+ B) = -\frac{57}{185}

Step-by-step explanation:

<u><em>the complete answer in the attached document</em></u>

Part 1) we have

sin(A)=\frac{8}{17}

cos(B)=\frac{12}{13}

Determine cos (A+B)

we know that

cos(A + B) = cos(A) cos(B)-sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{8}{17})^2=1

cos^2(A)+\frac{64}{289}=1

cos^2(A)=1-\frac{64}{289}

cos^2(A)=\frac{225}{289}

cos(A)=\pm\frac{15}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{15}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{12}{13})^2=1

sin^2(B)+\frac{144}{169}=1

sin^2(B)=1-\frac{144}{169}

sin^2(B)=\frac{25}{169}

sin(B)=\pm\frac{25}{169}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{5}{13}

step 3

Find cos(A+B)

substitute in the formula

cos(A + B) = \frac{15}{17} \frac{12}{13}-\frac{8}{17}\frac{5}{13}

cos(A + B) = \frac{180}{221}-\frac{40}{221}

cos(A + B) = \frac{140}{221}

Part 2) we have

sin(A)=\frac{3}{5}

cos(B)=\frac{12}{37}

Determine cos (A-B)

we know that

cos(A - B) = cos(A) cos(B)+sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{3}{5})^2=1

cos^2(A)+\frac{9}{25}=1

cos^2(A)=1-\frac{9}{25}

cos^2(A)=\frac{16}{25}

cos(A)=\pm\frac{4}{5}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{4}{5}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{12}{37})^2=1

sin^2(B)+\frac{144}{1,369}=1

sin^2(B)=1-\frac{144}{1,369}

sin^2(B)=\frac{1,225}{1,369}

sin(B)=\pm\frac{35}{37}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{35}{37}

step 3

Find cos(A-B)

substitute in the formula

cos(A - B) = \frac{4}{5} \frac{12}{37}+\frac{3}{5} \frac{35}{37}

cos(A - B) = \frac{48}{185}+\frac{105}{185}

cos(A - B) = \frac{153}{185}

Part 3) we have

sin(A)=\frac{15}{17}

cos(B)=\frac{3}{5}

Determine cos (A-B)

we know that

cos(A - B) = cos(A) cos(B)+sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{15}{17})^2=1

cos^2(A)+\frac{225}{289}=1

cos^2(A)=1-\frac{225}{289}

cos^2(A)=\frac{64}{289}

cos(A)=\pm\frac{8}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{8}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{3}{5})^2=1

sin^2(B)+\frac{9}{25}=1

sin^2(B)=1-\frac{9}{25}

sin^2(B)=\frac{16}{25}

sin(B)=\pm\frac{4}{5}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{4}{5}

step 3

Find cos(A-B)

substitute in the formula

cos(A - B) = \frac{8}{17} \frac{3}{5}+\frac{15}{17} \frac{4}{5}

cos(A - B) = \frac{24}{85}+\frac{60}{85}

cos(A - B) = \frac{84}{85}

Part 4) we have

sin(A)=\frac{15}{17}        

cos(B)=\frac{3}{5}

Determine cos (A+B)

we know that    

cos(A + B) = cos(A) cos(B)-sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{15}{17})^2=1

cos^2(A)+\frac{225}{289}=1

cos^2(A)=1-\frac{225}{289}      

cos^2(A)=\frac{64}{289}

cos(A)=\pm\frac{8}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{8}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{3}{5})^2=1

sin^2(B)+\frac{9}{25}=1

sin^2(B)=1-\frac{9}{25}

sin^2(B)=\frac{16}{25}

sin(B)=\pm\frac{4}{5}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{4}{5}

step 3

Find cos(A+B)

substitute in the formula    

cos(A + B) = \frac{8}{17} \frac{3}{5}-\frac{15}{17} \frac{4}{5}

cos(A + B) = \frac{24}{85}-\frac{60}{85}

cos(A + B) = -\frac{36}{85}

Download odt
4 0
3 years ago
20 to 16 percent error
Allisa [31]
The answer for this is 25 percent
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3 years ago
What is: <br> 5+5(5-5•5)-5 = ?
Alik [6]

Answer:

-100

Step-by-step explanation:

5 0
3 years ago
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