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2 years ago

Tell whether the angles are complementary or supplementary. Then find the value of x.

Mathematics

1 answer:

oee [108]2 years ago

6 0

Answer:

Supplementary angles,

x = 50

Step-by-step explanation:

These angles are supplementary, meaning that they add up to form a 180-degree angle or a straight line.

Knowing this, now form and solve the equation,

76 + 2x + 4 = 180

Combine like terms,

76 + 2x + 4 = 180

80 + 2x = 180

Inverse operations,

80 + 2x = 180

-80 -80

2x = 100

/2 /2

x = 50

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Indicate in standard form the equation of the line passing through the given point and having the given slope. r(4, 0), m = 5

Brrunno [24]

Y = mx + b....m is the slope and b is the y intercept
slope(m) = 5...(4,0)...x = 4 and y = 0
now we sub our info into y = mx + b...we r looking for b, the y intercept
0 = 5(4) + b
0 = 20 + b
-20 = b
so ur equation is y = 20x - 20...but we need it in standard form
Ax + By = C

y = 20x - 20....subtract 20x from both sides
-20x + y = -20...multiply by -1 to make x positive
20x - y = 20 <== standard form

5 0

2 years ago

What is the perimeter of the triangle shown on the coordinate plane,to the nearest tenth of a unit ?

ollegr [7]

Answer:

25.6 units

Step-by-step explanation:

From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).

First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:

$d=\sqrt{(x_2-x_1)^{2} +(y_2-y_1)^{2}}$

where

$(x_1,y_1)$ are the coordinates of the first point

$(x_2,y_2)$ are the coordinates of the second point

- For AB:

$d=\sqrt{[1-(-5)]^{2}+(4-4)^2}$

$d=\sqrt{(1+5)^{2}+(0)^2}$

$d=\sqrt{(6)^{2}}$

$d=6$

- For BC:

$d=\sqrt{(3-1)^{2} +(-4-4)^{2}}$

$d=\sqrt{(2)^{2} +(-8)^{2}}$

$d=\sqrt{4+64}$

$d=\sqrt{68}$

$d=8.24$

- For AC:

$d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}$

$d=\sqrt{(3+5)^{2} +(-8)^{2}}$

$d=\sqrt{(8)^{2} +64}$

$d=\sqrt{64+64}$

$d=\sqrt{128}$

$d=11.31$

Next, now that we have our lengths, we can add them to find the perimeter of our triangle:

$p=AB+BC+AC$

$p=6+8.24+11.31$

$p=25.55$

$p=25.6$

We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.

6 0

3 years ago

Identify the terms themselves for the following expression. (separate ur terms with a comma)

Zinaida [17]

3a, 5b, -6y, 10 hope this helps

7 0

2 years ago

Read 2 more answers

Jim has only dimes and nickels in his coin bank. He has 37 coins with a total value of $2.65.

guajiro [1.7K]

Answer:

Step-by-step explanation:

Let x represent the number of dimes and y represent the number of nickels. The total number of coins is 37; this gives us the equation

x+y = 37

Each dime is worth ten cents, or 0.10, and each nickel is worth five cents, or 0.05. The total amount of money is given by

0.10x+0.05y = 2.65

This gives us the system

$\left \{ {{x+y=37} \atop {0.10x+0.05y=2.65}} \right.$

To solve this, we will use substitution. We will isolate x in the first equation:

x+y=37

Subtract y from each side:

x+y-y = 37-y

x = 37-y

Substitute this into the second equation:

0.10(37-y)+0.05y = 2.65

Using the distributive property,

0.10(37)-0.10(y)+0.05y = 2.65

3.70-0.10y+0.05y = 2.65

Combining like terms,

3.70-0.05y = 2.65

Subtract 3.70 from each side,

3.70-0.05y-3.70 = 2.65-3.70

-0.05y = -1.05

Divide both sides by -0.05:

-0.05y/-0.05 = -1.05/-0.05

y = 21

There were 21 nickels.

4 0

2 years ago

Read 2 more answers

A house was valued at $100,000 in the year 1987. The value appreciated to $165,000 by the year 2002.

iris [78.8K]

Answer:

A) The annual multiplier was 1.0339; the annual increase was 0.0339 of the value.

B) 3.39% per year

C) $182,000

Step-by-step explanation:

A) Let's let t represent years since 1987. Then we can fill in the numbers and solve for r.

165000 = 100000(1 +r)^15

1.65^(1/15) = 1 +r . . . . . divide by 100,000; take the 15th root

1.03394855265 -1 = r ≈ 0.0339

The value was multiplied by about 1.0339 each year.

B) The value increased by about 3.39% per year.

C) S = $100,000(1.03394855265)^18 ≈ $182,000

8 0

3 years ago