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mariarad [96]
3 years ago
8

NEEEEEEEEDDDDDDDDS HELLLPLLPP

Mathematics
1 answer:
Morgarella [4.7K]3 years ago
3 0
We know that angle2+angle3+90=180 (since we see the right angle in the corner and all internal angles add up to 180) so

angle3=40
subsitute
angle2+40+90=180
angle2+130=180
subtract 130
angle2=50
answer is 50 degrees
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Im assuming that's a regular octagon. The sum of the angles in an octagon is 1080, so that means each individual angle is 135.

m<1 = 135.

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4 0
4 years ago
tiyleah has a cousin fantasia in france . both families recently bought new cars and the two girls are comparing how fuel effici
Kay [80]

Answer:

Fantasia's family's car is more fuel-efficient.

Step-by-step explanation:

Given that, Tiyleah's family's car is getting 40 miles per gallon, and Fantasia's family's car uses 7 liters per 150 km.

The comparison can be done by observing the distance traveled by the cars in an equal amount of fuel. The car which covers more distance in an equal amount of fuel is more fuel-efficient.

First, convert the given units to have the same unit of measurement,

As 1 gallon =3.7854 liters and 1 mile = 1.609344 km

So, 40 miles =40 x 1.61=64.40 km

Now, The distance covered by Tiyleah's family's car is 64.40 km in 3.7854 liters.

So, the distance covered by Tiyleah's family's car in 1 liter of fuel is \frac{64.40}{3.7854}=17.01 km

In a similar way,

As Fantasia's family's car uses 7 liters per 150 km.

So, the distance traveled by Fantasia's family's car in 1 liter of fuel =150/7=21.43 km

Observe that, in 1 liter of fuel, Tyleah's family's car covered 17.01 km while Fantasia's family's car covered 21.43 km which is more.

Hence, Fantasia's family's car is more fuel-efficient.

5 0
4 years ago
Find an equation for the nth term of the sequence.<br><br> -3, -12, -48, -192, ...
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Answer:

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3 years ago
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5 0
3 years ago
Can u help me pls they are exponents with multiplication and division
Ronch [10]
When there is a multiplication involving exponents, we need to evaluate the numbers and each variables separately. In this case, we have 5 and 4 as the numbers, and w and h for the variables.

When we multiply 5 and 4 we get:
5 \times 4 = 20

For w, we have {w}^{6}, {w}^{4}, w. While we are multiplying the w's, we need to add the exponents together to get the final answer: {w}^{6} \times {w}^{4} \times w = {w}^{11}

For h's, there are {h}^{2}, {h}^{5}, {h}^{4}. When we apply the same thing to h's, we get {h}^{2} \times {h}^{5} \times {h}^{4} = {h}^{11}

For the final step, we need to write all of these together: 20 \times {w}^{11} \times {h}^{11} \: \: or \: \: 20 {w}^{11} {h}^{11} \: \: or \: \: 20 {(wh)}^{11}
3 0
4 years ago
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