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3 years ago

A box of baseball cards is $25.00 with %15 off. What is the sale price of the box of cards? In other words, how much is it after

Mathematics

1 answer:

yan [13]3 years ago

8 0

Answer:

21.25

Step-by-step explanation:

you would multiply 25 by 0.15 to get the percent being taken off then subtract that from your original price

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What is the absolute value of +25 A. -25 B. 25

bogdanovich [222]

B is the correct answer. The absolute value of something will always be positive. ALWAYS.

3 0

3 years ago

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gtnhenbr [62]

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3 years ago

What is the value of x? Enter your answer in the box. °

Gnom [1K]

Answer:

The answer to your question is: 111°

Step-by-step explanation:

The sum of the interior angles of a polygon is (n - 2) 180°

This polygon has 7 sides

Then n = 7

Process

120 + 158 + 126 + 125 + 121 + 139 + x = (7 - 2)180

120 + 158 + 126 + 125 + 121 + 139 + x = 5(180)

789 + x = 900

x = 900 -789

x = 111°

5 0

3 years ago

What is the missing length.

nasty-shy [4]

Answer:

$KS = 12$

Step-by-step explanation:

Given that ∆KLM ~ ∆RSK, $\frac{KL}{KR} = \frac{KM}{KS}$ (similarity theorem)

KL = 65

KR = 65 - 52 = 13

KM = 60

KS = ?

$\frac{65}{13} = \frac{60}{KS}$

Cross multiply

$65*KS = 60*13$

$65*KS = 780$

$\frac{65*KS}{65} = \frac{780}{65}$

$KS = 12$

4 0

3 years ago

Triangle ABC has AB=5, BC=7, and AC=9. D is on AC with BD=5. Find the length of DC

e-lub [12.9K]

Answer:

$DC=\frac{8}{3}\ units$

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The triangle ABD is an isosceles triangle

because

AB=BD

The segment BM is a perpendicular bisector segment AD

In the right triangle ABM

Applying the Pythagorean Theorem

$BM^2=AB^2-AM^2$

we have

$AB=5\ units\\AM=x\ units$

substitute

$BM^2=5^2-x^2$

$BM^2=25-x^2$ -----> equation A

In the right triangle BMC

Applying the Pythagorean Theorem

$BM^2=BC^2-MC^2$

we have

$BC=7\ units\\MC=AC-AM=(9-x)\ units$

substitute

$BM^2=7^2-(9-x)^2$

$BM^2=49-(81-18x+x^2)$

$BM^2=49-81+18x-x^2$

$BM^2=-x^2+18x-32$ ----> equation B

equate equation A and equation B

$-x^2+18x-32=25-x^2$

solve for x

$18x=25+32\\18x=57\\\\x=\frac{57}{18}$

Simplify

$x=\frac{19}{6}$

Find the length of DC

$DC=AC-2x$

substitute the given values

$DC=9-2(\frac{19}{6})$

$DC=9-\frac{19}{3}\\\\DC=\frac{8}{3}\ units$

8 0

3 years ago