How to graph -x+y=-6 , 3x+5y=2

1. Mark charges $10 an hour to babysit with an additional flat fee of $20 for each babysitting job. He wants to purchase a new g

quester [9]

Answer:

Step-by-step explanation:

A.123.50≤20+10(x)

103.50≤10x

10.35 hours at least to equal the 123.50.

10.35 hours *10 an hour is 103.50+20 flat fee=123.50

B. 123.50≤15(x)

8.24 hours or more. 15.00 an hour at 8.24 hours = 123.60

C. He will work less hours if he charges more and does not charge the flat fee.

6 0

3 years ago

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The numbers in this chart represent the weights in kilograms of five

Nitella [24]

John’s is the outlier

3 0

3 years ago

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The measure of an angle is three times the measure of its supplementary angle. What is the measure of each angle?

madreJ [45]

Answer:

45°

135°

Step-by-step explanation:

Let the measure of one angle be x

So, measure of its supplementary angle = 3x

$\implies \: x + 3x = 180 \degree \\ \\ \implies \: 4x = 180 \degree \\ \\\implies \: x = \frac{ 180 \degree}{4} \\ \\\implies \: x = 45 \degree \\ \\ \implies \: 3x = 135 \degree$

5 0

2 years ago

In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?

snow_lady [41]

If ΔACB is an isosceles triangle, then ∠A ≅ ∠B and AC ≅ CB

Since ∠C = 120° and ∠A + ∠B + ∠C = 180°, then ∠A = 30° and ∠B = 30°

Next, look at ΔADB. ∠A + ∠D + ∠B = 180°, so ∠A + 90° + 30° = 180° ⇒ ∠A = 30°

Now look at ΔADC. Since ∠A = 30° in ΔACB, and ∠A = 60° in ΔADB, then ∠A = 30° in ΔADC <em>per angle addition postulate.</em>

Now that we have shown that ΔADB and ΔADC are 30-60-90 triangles, we can use that formula to calculate the side lengths.

CD = 4 cm (given) so AC = 2(4 cm) = 8 cm

Since AC ≅ BC, then BC = 8 cm. Therefore, BD = 4 + 8 = 12 <em>by segment addition postulate.</em>

Lastly, look at ΔBHD. Since ∠B = 30° and ∠H = 90°, then ∠D = 60°. So, ΔBHD is also a 30-60-90 triangle.

BD = 12 cm, so HD = $\frac{12}{2}cm$ = 6 cm

Answer: 6 cm

8 0

3 years ago

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The center of a hyperbola is (−4,3) , and one vertex is (−4,7) . The slope of one of the asymptotes is 2.

Monica [59]

Answer:

The answer to your question is below

Step-by-step explanation:

C (-4, 3)

V (-4, 7)

asymptotes = 2 = $\frac{b}{a}$

- This is a vertical hyperbola, the equation is

$\frac{(y - k)^{2} }{a^{2} } + \frac{(x - h)^{2} }{b^{2} } = 1$

slope = 2

a is the distance from the center to the vertex = 4

b = 2(4) = 8

$\frac{(y - 3)^{2} }{4^{2} } + \frac{(x + 4)^{2} }{8^{2} } = 1$

$\frac{(y - 3)^{2} }{16} + \frac{(x + 4)^{2} }{64} = 1$

7 0

3 years ago

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