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Anit [1.1K]
2 years ago
15

Help please i know its alot but i am in danger of failing for the year if i fail this test :(

Mathematics
1 answer:
Lena [83]2 years ago
3 0

Answer:

(20.)  D is the correct answer

(19.) A is correct

(18.) A is correct

(17.) D is correct

(15.) A is correct

Step-by-step explanation:

hope this helps u ;)

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Find the slope of the line that passes through (2, 4) and (9, 9).
Verdich [7]
M = 5/7


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8 0
3 years ago
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Which equation can be used to solve for m, the greater integer? m(m – 3) = 108 m(m + 3) = 108 (m + 3)(m – 3) = 108 (m – 12)(m –
nata0808 [166]

Answer:

m(m-3)=108

Step-by-step explanation:

Complete question below:

Two positive integers are 3 units apart on a number line. Their product is 108.

Which equation can be used to solve for m, the greater integer?

m(m – 3) = 108

m(m + 3) = 108

(m + 3)(m – 3) = 108

(m – 12)(m – 9) = 108

Solution

On the number line,

Let

m= larger integer

The integers are 3 numbers apart on the number line, so

m-3=smaller integer

The product (×) of the larger and smaller integers=108

(m)*(m-3)=108

m(m-3)=108

Therefore, the equation that can be used to solve for m, the larger integer is:

 m(m – 3) = 108 

5 0
3 years ago
How would you find the missing x and y values if you don’t have a model or equation to go by?
mamaluj [8]
You can find the slope and y-intercept, and make an equation in slope- intercept form and them just plug in the other x values in the equation.

3 0
3 years ago
The mean hourly salary of the 10 employees at a fast­food restaurant is $8.25. One of the employees earning $6.50 an hour leaves
IgorC [24]
You would add $8.25 + $5.50, which equals $13.75. Then you would divide that number by the number of employees currently working at the fast food restaurant; $13.75 / 2 = $6.875. So, $6.88 would be the new mean salary of the employees.
8 0
3 years ago
In this triangle, the product of sin B and tan C is<br> and the product of sin Cand tan B is
Ipatiy [6.2K]

Answer:

\frac{c}{a} and \frac{b}{a}

Step-by-step explanation:

sinB = \frac{opposite}{hypotenuse} = \frac{AC}{BC} = \frac{b}{a}

tanC = \frac{opposite}{adjacent} = \frac{AB}{AC} = \frac{c}{b}

Thus

sinB tanC = \frac{b}{a} × \frac{c}{b} ( cancel b on numerator/ denominator )

                 = \frac{c}{a}

---------------------------------------------------------------------------

sinC = \frac{opposite}{hypotenuse} = \frac{AB}{BC} = \frac{c}{a}

tanB = \frac{opposite}{adjacent} = \frac{AC}{AB} = \frac{b}{c}

Thus

sinC tanB = \frac{c}{a} × \frac{b}{c} ( cancel c on numerator/ denominator )

                 = \frac{b}{a}

7 0
2 years ago
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