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polet [3.4K]
3 years ago
5

42 = 7 ( x + 5 ) solve for x

Mathematics
2 answers:
PSYCHO15rus [73]3 years ago
7 0

Answer:

1

Step-by-step explanation:

1. distribute 7 to x and 5.

42 = 7x + 35

2. subtract 35 on both sides.

7 = 7x

3. Divide by 7 for both sides.

1 = x

Olegator [25]3 years ago
4 0

Answer:

x = 1

Step-by-step explanation:

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The unit rate is a ratio of change in y and into the change on x. It can also be the rise to run or the slope.
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Jackie is selling smoothies at a school fair. She starts the day with $15 in her cash box to provide change to her customers. If
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A city had population 67,255 on january 1, 2000, and its population has been increasing by 2935 people each year since then. A l
irina1246 [14]

We are given: On january 1, 2000 initial population   = 67,255.

Number of people increase each year = 2935 people.

Therefore, 67,255 would be fix value and 2935 is the rate at which population increase.

Let us assume there would be t number of years after year 2000 and population P after t years is taken by function P(t).

So, we can setup an equation as

Total population after t years = Number of t years * rate of increase of population + fix given population.

In terms of function it can be written as

P(t) = t * 2935 + 67255.

Therefore, final function would be

P(t) = 2935t +67255.

So, the correct option is d.P(t) = 67255 + 2935t.

4 0
3 years ago
Read 2 more answers
A) 7xy?<br> Degree and number of terms
jolli1 [7]

Answer:

1 and 1

Step-by-step explanation:

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8 0
3 years ago
Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137
Nata [24]

Answer:

<em>Camera 2nd has to cover the maximum angle, i.e. </em>78.70^\circ.

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle \triangle ABC with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles \angle A, \angle B, \angle C

As per Law of cosine:

cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}

Putting the values of a,b and c to find out angles \angle A, \angle B, \angle C.

cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ

cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ

cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ

<em>Camera 2nd has to cover the maximum angle</em>, i.e. 78.70^\circ.

6 0
3 years ago
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