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Bezzdna [24]
4 years ago
7

Write an equivalent expression for -3t+4q+5t-7q

Mathematics
1 answer:
steposvetlana [31]4 years ago
6 0

Answer:

The equivalent expression can be given as:

⇒  2t-3q

Step-by-step explanation:

Given expression:

-3t+4q+5t-7q

To write an equivalent expression:

Solution:

In order to write an equivalent expression, we will combine the like terms. By like terms, we mean the terms with the same variable constant.

For example : 5x and -3x  are like terms.

We have:

-3t+4q+5t-7q

Combining like terms :

⇒ -3t+5t+4q-7q

⇒  2t-3q

Thus, the equivalent expression can be given as:

⇒  2t-3q

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If a and b are two angles in standard position in Quadrant I, find cos(a+b) for the given function values. sin a=4/5 and cos b=5
vladimir1956 [14]
First let's find the angles a and b.

 We have then:
 sin a = 4/5
 a = Asin (4/5)
 a = 53.13 degrees.

 cos b = 5/13
 b = Acos5 / 13
 b = 67.38 degrees.

 We now calculate cos (a + b). To do this, we replace the previously found values:
 cos ((53.13) + (67.38)) = - 0.507688738
 Answer: 
 -0.507688738
 Note: there is another way to solve the problem using trigonometric identities.
4 0
4 years ago
Read 2 more answers
Y=4/5(188)+113.75 solve for y
VARVARA [1.3K]

Answer:

264.15

Step-by-step explanation:

4/5(188)+113.75

=0.8(188)+113.75

=150.4+113.75

=264.15

3 0
3 years ago
Buddy ran the same distance each day for 5 days.
bagirrra123 [75]

Answer:

A.

One expression you can do is the total amount of distance Buddy runs in a 5 day period, as a function of the amount of miles he runs in a day.

d(x) = y = 5x

d(x) being total amount of miles buddy runs in 5 days. x being the amount of miles he runs in a day.

B.

Now if he ran 10 miles a day, just substitute 10 for x.

d(10) = y = 5(10) = 50

8 0
3 years ago
PLEASE HELP I NEED THE CORRECT ANSWER FAST
Drupady [299]
22 degrees is the answer
3 0
3 years ago
Read 2 more answers
Suppose that you were not given the sample mean and sample standard deviation and instead you were given a list of data for the
Troyanec [42]

Answer:

So, the sample mean is 31.3.

So, the sample standard deviation is 6.98.

Step-by-step explanation:

We have a list of data for the speeds (in miles per hour) of the 20 vehicles. So, N=20.

We calculate the sample mean :

\mu=\frac{19 +19 +22 +24 +25 +27 +28+ 37 +35 +30+ 37+ 36+ 39+ 40+ 43+ 30+ 31+ 36+ 33+ 35}{20}\\\\\mu=\frac{626}{20}\\\\\mu=31.3

So, the sample mean is 31.3.

We use the formula for a sample standard deviation:

\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}

Now, we calculate the sum

\sum_{i=1}^{20}(x_i-31.3)^2=(19-31.3)^2+(19-31.3)^2+(22-31.3)^2+(24-31.3)^2+(25-31.3)^2+(27-31.3)^2+(28-31.3)^2+(37-31.3)^2+(35-31.3)^2+(30-31.3)^2+(37-31.3)^2+(36-31.3)^2+(39-31.3)^2+(40-31.3)^2+(43-31.3)^2+(30-31.3)^2+(31-31.3)^2+(36-31.3)^2+(33-31.3)^2+(35-31.3)^2\\\\\sum_{i=1}^{20}(x_i-31.3})^2=926.2\\

Therefore, we get

\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}\\\\\sigma=\sqrt{\frac{1}{19}\cdot926.2}\\\\\sigma=6.98

So, the sample standard deviation is 6.98.

3 0
3 years ago
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