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Pavlova-9 [17]
3 years ago
11

What equation is graphed in this figure?

Mathematics
1 answer:
Vilka [71]3 years ago
4 0
I believe it’s the last answer the fourth one
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How much would 3 yd of lumber cost if each foot costs $3.75
Shtirlitz [24]

Answer:

$33.75

Step-by-step explanation:

3 ft are in a yd so 9 ft are in 3 yd. Then you multiply 9 by 3.75 which brings you to the answer $33.75.

5 0
3 years ago
Can u plz answer this it’s for a test !! 15&16 PLZ
ss7ja [257]

Answer:

15. 6647.61

16. 553

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Are these angles complementary, supplementary, or neither?
SashulF [63]
I would say complementary because the angle degrees are the same
4 0
3 years ago
Fill in the blank with a constant, so that the resulting expression can be factored as the product of two linear expressions: 2a
Vitek1552 [10]

Answer:

2ab - 6a + 5b - 15

Step-by-step explanation:

Given

2ab - 6a + 5b + \_

Required

Fill in the gap to produce the product of linear expressions

2ab - 6a + 5b + \_

Split to 2

(2ab - 6a) + (5b + \_)

Factorize the first bracket

2a(b - 3) + (5b + \_)

Represent the _ with X

2a(b - 3) + (5b + X)

Factorize the second bracket

2a(b - 3) + 5(b + \frac{X}{5})

To result in a linear expression, then the following condition must be satisfied;

b - 3 = b + \frac{X}{5}

Subtract b from both sides

b - b- 3 = b - b+ \frac{X}{5}

- 3 = \frac{X}{5}

Multiply both sides by 5

- 3 * 5 = \frac{X}{5} * 5

X = -15

Substitute -15 for X in 2a(b - 3) + 5(b + \frac{X}{5})

2a(b - 3) + 5(b + \frac{-15}{5})

2a(b - 3) + 5(b - \frac{15}{5})

2a(b - 3) + 5(b - 3)

(2a + 5)(b - 3)

The two linear expressions are (2a+ 5) and (b - 3)

Their product will result in 2ab - 6a + 5b - 15

<em>Hence, the constant is -15</em>

3 0
3 years ago
An airplane travels about 1,500 Km in a straight line between Toronto and Winnipeg. However, the plane must climb to 10,000 m du
Anna007 [38]

Answer:416

Step-by-step explanation: Based on the given problem above, it requires a trigonometric solution. So here it goes:

The climb distance is:200 * Sec [ArcTan [10/200]] = 10√ 401 km

and the decent distance is:300 * Sec [ArcTan [10/300]] = 10√901 km

So add the given results above with 500 km and this will be the additional distance that plane moves through the air.

The answer would be in the unit meters.

((10√401 +10√901−500)∗1000)=416 meters

Hope this is the answer that you are looking for.

7 0
2 years ago
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