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2 years ago

11

ASAP GIVING GOOD AMOUNT OF POINTS. I cannot fail my class or I will be sent to school again

1 answer:

svlad2 [7]2 years ago

6 0

Answer

- 3/2

Step-by-step explanation:

Hope this helps:) if you need anynore help lmk!

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a piece of string is 3 yards long. How many 1 and a quarter yard long pieces can Julie cut from the string?

Georgia [21]

This question is actually a lot simpler than it seems

for a quarter think about money - a quarter is 25 cents or 1/4 of a dollar - so 1 and a quarter yard in decimal form is equivalent to 1.25

now you need to find how many 1 and a quarter yards are in 3 yards - to do this you simply divide 3 by 1.25

3/1.25 = 2.4

so julie can cut 2.4 1 and a quarter long pieces from the string

if you have any questions about how i answered this question please let me know:)

3 0

3 years ago

Read 2 more answers

7x + 8 + x=<br><br>plsss i need help i will cashapp u 5 dollars help<br> mee

MA_775_DIABLO [31]

I think the answer
= 8x + 8

6 0

3 years ago

Read 2 more answers

What is the answer to -3x = 4.2 ?

hammer [34]

Answer:

$- 3x = 4.2 \\ - x = \frac{4.2}{3} \\ - x = 1.4 \\ x = - 1.4 \\ thank \: you$

4 0

2 years ago

Please answer quickly who ever answers the question correctly will get a brainlyest + 100 points.

bazaltina [42]

Answer:

Step-by-step explanation:

The perimeter is all the side lengths added together

In this case, the

perimeter = x + (x+4) + (x+6) = 3x + 10

The correct answer is the first choice

3 0

2 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

so

<em>In the right triangle ABM</em>

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

<em>In the right triangle BMC</em>

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

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

<em>Find the length of DC</em>

$DC=AC-2x$

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