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3 years ago

Question 1 - Part A

Mathematics

1 answer:

Vikki [24]3 years ago

3 0

Answer:

A and C

Step-by-step explanation:

Part A

150+30x, x being the amount of visits she takes

Part C

150+30(3)

150+90

240

$240

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If you reflect triangle MOV across the x-axis, what is the coordinate of O in the image?

Zina [86]

Hi,

The answer you are looking for is 2. (-4,2)

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3 years ago

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Translate to an equation and solve: The sum of two-fifths and f is one-half.<br> Provide<br> f=

SCORPION-xisa [38]

Answer:

9/10

1/2+2/5

Since these fractions have different denominators, we need to find the least common multiple of the denominators.The least common multiple of 2 and 5 is 10, so we need to multiply to make each of the denominators = 10

1/2 ∗ 5/5 = 5/10

5/2∗ 2/2= 4/10

Since these fractions have the same denominator, we can just add the numerators

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2 years ago

A carnival charges $8 to enter and $3 per ride. Fill in the table of values for the total cost for a person to enter and go on 0

grigory [225]

Answer:

Look at the picture

Step-by-step explanation:

For 0 rise it would just be eight dollars

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2 years ago

50 POINTS!!! In rectangle ABCD, AB = 6 cm, BC = 8 cm, and DE = DF. The area of triangle DEF is one-fourth the area of rectangle

aalyn [17]

Answer:

$EF=4\sqrt{3}$

Step-by-step explanation:

In rectangle ABCD, AB = 6, BC = 8, and DE = DF.

ΔDEF is one-fourth the area of rectangle ABCD.

We want to determine the length of EF.

First, we can find the area of the rectangle. Since the length AB and width BC measures 6 by 8, the area of the rectangle is:

$A_{\text{rect}}=8(6)=48\text{ cm}^2$

The area of the triangle is 1/4 of this. Therefore:

$\displaystyle A_{\text{tri}}=\frac{1}{4}(48)=12\text{ cm}^2$

The area of a triangle is half of its base times its height. The base and height of the triangle is DE and DF. Therefore:

$\displaystyle 12=\frac{1}{2}(DE)(DF)$

Since DE = DF:

$24=DF^2$

Thus:

$DF=\sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}=DE$

Since ABCD is a rectangle, ∠D is a right angle. Then by the Pythagorean Theorem:

$(DE)^2+(DF)^2=(EF)^2$

Therefore:

$(2\sqrt6)^2+(2\sqrt6)^2=EF^2$

Square:

$24+24=EF^2$

Add:

$EF^2=48$

And finally, we can take the square root of both sides:

$EF=\sqrt{48}=\sqrt{16\cdot 3}=4\sqrt{3}$

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3 years ago

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Zielflug [23.3K]

Answer:

-----

Step-by-step explanation:

3 5 15

-- * -- == ---

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2 years ago