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3 years ago

1. In AFKR, which side is included between

Mathematics

1 answer:

Igoryamba3 years ago

4 0

Answer:

(B.)k

Step-by-step explanation:

in between F and R is K.

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Can someone help w number 4

strojnjashka [21]

Answer:

angle DFG = 49 degrees

angle JKL = 41 degrees

Step-by-step explanation:

When angles are complementary with each other, it means that if you add both of the angles up, it adds up to 90 degrees.

In this question, you would have to add up angle DFG and angle JKL and find the x that makes the equation equal to 90 degrees.

angle DFG = x + 5

angle JKL = x - 3

(x + 5) + (x - 3) = 90

2x + 2 = 90

2x = 90 - 2

2x = 88

x = 44

But since we have to find out the angle measures, we have to the "x = 44" with the x's in the DFG and JKL angles.

DFG = (44) + 5 = 49

JKL = (44) - 3 = 41

7 0

3 years ago

Please answer correctly

Afina-wow [57]

Answer:

2nd and 4th options

Step-by-step explanation:

both answers begin with a fixed point drawn with a straight line through a point that continues on by showing an arrow

3 0

2 years ago

Read 2 more answers

A 120-watt light bulb uses about 0.1 kilowatt of electricity per hour. If electricity costs $0.20 per kilowatt hour, how much do

saveliy_v [14]

There is no such thing as a "kilowatt per hour". If that's actually what the question says, then it's a defective question, and you should put it away before it makes you any more confused. A 120 watt light bulb uses exactly 0.12 kilowatt when it's turned on. In one hour, it uses (0.12 kilowatt) x (1 hour) = 0.12 kilowatt-hour of energy. If energy costs $0.20 per kilowatt-hour, then the cost is (0.12) x (0.20) = 2.4 cents. (0.024 dollar)

8 0

3 years ago

Expand using the properties and rules for logarithms

malfutka [58]

Consider expression $\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).$

1. Use property

$\log_a\dfrac{b}{c}=\log_ab-\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.$

2. Use property

$\log_abc=\log_ab+\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2.$

3. Use property

$\log_ab^k=k\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2.$

4. Use property

$\log_{a^k}b=\dfrac{1}{k}\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.$

Answer: correct option is B.

7 0

3 years ago

What is the slope of the line that passes through (2 -- 3) and (-2, 10)

VMariaS [17]

Answer:

13 / -4

Step-by-step explanation:

7 0

2 years ago