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fiasKO [112]
3 years ago
15

The Robinson family traveled a total of 240 miles in 6 hours on their trip to visit relatives. If they travel at a constant, pro

portional rate, write an equation relating the number of miles traveled mto the number of hours h driven. Then use your equation to figure out how many miles they would travel in 13 hours.
Mathematics
1 answer:
ivanzaharov [21]3 years ago
6 0

Answer:520

Step-by-step explanation:

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Find the complementary angle of the following <br>1)69°. 2)39°. 3)17°. 4)40°​
Luden [163]

Answer:

1.)21°

2.)51°

3.)73°

4.)50°

Step-by-step explanation:

1.) 90° - 69°

= 21°

2.) 90° - 39°

= 51°

3.) 90° - 17°

= 73°

4.) 90° - 40°

= 50°

3 0
3 years ago
Can someone help me with this problem<br> I don't understand.
Ivahew [28]
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Because you are distributing the 3 to 2/5x and 7/5y
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3 years ago
How to solve transformation problems in math
Molodets [167]
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4 0
4 years ago
Write one digit on each side of 10 to make a four digit multiple of 72. How many different solutions does this problem have? The
Viefleur [7K]

If we write 4 to each side of 10, we will get a four digit number 4104, which is a multiple of 72.

To find out other possible solutions, first start with 1 in front of 10.

Now, divide 110 by 72.

The remainder is 38.

We are going to put one more digit after 10.

So, let us check if there is any three digit number whose first two digits are 38 and which is a multiple of 72.

There is no such number since 5 × 72 = 360 and 6 × 72 = 432.

So, there is no four digit number whose first three digits are 110 and which is a multiple of 72.

If we try like this, we can find 4104 is the only solution.

5 0
4 years ago
CosA + cosB - cosC = -1 + 4cosA/2 cosB/2 sinC/2
Lubov Fominskaja [6]

Answer:

Step-by-step explanation:

(cos A+ cos B)-cos C

=2cos \frac{A+B}{2}cos \frac{A-B}{2}-cos C~~~...(1)\\A+B+C=180\\A+B=180-C\\\frac{A+B}{2}=90-\frac{C}{2}\\cos \frac{A+B}{2}=cos(90-\frac{C}{2})=sin \frac{C}{2}\\cos C=1-2sin^2\frac{C}{2}\\(1)=2 sin \frac{C}{2} cos \frac{A-B}{2}-1+2sin^2\frac{C}2}\\=2sin\frac{C}{2}[cos \frac{A-B}{2}+sin \frac{C}{2}]-1~~~...(2)\\\\now~again~A+B+C=180\\C=180-(A+B)\\sin\frac{C}{2}=sin(90-\frac{A+B}{2})=cos \frac{A+B}{2}\\(2)=2sin\frac {C}{2}[cos \frac{A-B}{2}+cos \frac{A+B}{2}]-1\\

=-1+2sin\frac{C}{2}*2cos \frac{\frac{A-B}{2} +\frac{A+B}{2} }{2} cos \frac{\frac{A-B}{2} -\frac{A+B}{2} }{2} \\=-1+4sin\frac{C}{2} cos \frac{A}{2} cos\frac{-B}{2} \\=-1+4 cos \frac{A}{2} cos \frac{B}{2} sin \frac{C}{2}\\(cos(-B)=cos B)

8 0
3 years ago
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