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tatyana61 [14]
3 years ago
7

Evaluate 5e + 5 -7f when e = 6 and f = 5

Mathematics
2 answers:
EastWind [94]3 years ago
8 0
5e+5-7f
5×6+5-7×5
30+5-35
35-35=0
Therefore the answer is 0
Dovator [93]3 years ago
3 0
5(6)+5-7(5)
30+5-35
35-35
0

Final answer: 0

Remember to perform multiplication before addition and subtraction according to the order of operations.
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Vadim26 [7]

Answer:

2. H

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kirza4 [7]

Answer:

The relationship of the number of months and the total amount paid is proportional for both Hanks and Lynn.

Step-by-step explanation:

Let us divide the equation into 2 parts.

Hanks

Hank paid $2000 up front when he bought the car and he pays $200 every month. Therefore, the total amount paid (y)  and the number of month (x)can be expressed as follows.

Let

x = number of month

y = total amount paid

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The relationship between amount paid and the number of months is proportional

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Mashutka [201]
The angle of elevation from the hiker to the park ranger is the same as the angle of depression from the ranger to the hiker, 29°.

The tangent function of an angle is the ratio of the opposite side (500') to the adjacent side (x) in the right triangle shown. That is
   tan(29°) = 500' / x

Multiplying the equation by x/tan(29°) gives
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3 years ago
Can you help me please??
dusya [7]

Answer:

y = \frac{4}{5}x+\frac{54}{5}

Step-by-step explanation:

Equation of a line has been given as,

y=\frac{4}{5}x+\frac{3}{5}

Here, slope of the line = \frac{4}{5}

y-intercept = \frac{3}{5}

"If the two lines are parallel, there slopes will be equal"

By this property slope of the parallel line to the given line will be equal.

Therefore, slope 'm' = \frac{4}{5}

Since, slope intercept form of a line is,

y = mx + b

Therefore, equation of the parallel line will be,

y = \frac{4}{5}x+b

Since, this line passes through a point (-6, 6),

6 = \frac{4}{5}(-6)+b

6 = -\frac{24}{5}+b

b = 6+\frac{24}{5}

b = \frac{30+24}{5}

b = \frac{54}{5}

Equation of the parallel line will be,

y = \frac{4}{5}x+\frac{54}{5}

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fiasKO [112]

Answer:

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In step 4 were proved that \triangle ABC\sim\triangle DEC.

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