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3 years ago

9

The vertices of triangle ABC are A(−5, 1), B(−2.5, 7), and C(0, 1). Find the new coordinates of the vertices after the given tra

nslation. (x, y) → (x + 3, y − 1)

1 answer:

morpeh [17]3 years ago

4 0

Answer:

(2,0) (0.5,6) (3,0)

Step-by-step explanation:

u know how it's done

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Solve c= 5/9(F-32) for F

guajiro [1.7K]

Answer:

F=9/5c+32

Step-by-step explanation:

c=5/9f-160/9

9c=5f-160

-5f=-160-9c

f=32+9/5c

f=9/5c+32

8 0

2 years ago

Solve 3(x + 2) > x.

Georgia [21]

3(x + 2) > x
3x + 6 > x
3x - 3x + 6 > x - 3x
6 > -2x
6/-2 < -2x/-2
-3< x
(When dividing by a negative, reverse the direction of the inequality side, reverse it EVERY time you divide by a negative)

ANSWER: -3 < x

5 0

3 years ago

2.1 If 5 + 3 = 0 and < 0, evaluate (without using a calculator):

Sergeeva-Olga [200]

1. 18

2.8

3.-1

<h3>What is expression?</h3>

An expression is a set of terms combined using the operations +, – , x or ,/.

Given:

1. 22 − 1 (4)

=22- 1*4

= 22-4

= 18

2. 2 + 2 (3)

=2+2*3

=2+6

=8

3. 1 − 2

= -1

Learn more about expression here:

brainly.com/question/14083225

#SPJ1

6 0

2 years ago

Mr. James accepted a new job with a starting annual salary of $32,500. He will receive an annual raise of $1,500 beginning his s

Amanda [17]

Answer:

He will earn $139,000 after 4 years

Step-by-step explanation:

32,500+34,000+35,500+37000= $139,000

7 0

3 years ago

A metallurgist has one alloy containing 34% copper another containing 48% copper. How many pounds of each alloy must he use to m

Zielflug [23.3K]

Answer:

<h2>36.14 pounds of 34% copper alloy and 9.86 pounds of 48% copper alloy</h2>

Step-by-step explanation:

First alloy contains 34% copper and the second alloy contains 48% alloy.

We wish to make 46 pounds of a third alloy containing 37% copper.

Let the weight of first alloy used be $x$ in pounds and the weight of second alloy used be $y$ in pounds.

Total weight = $46\text{ }pounds=x+y$ $-(i)$

Total weight of copper = $37\%\text{ of 46 pounds = }34\%\text{ of }x\text{ pounds + }48\%\text{ of }y\text{ pounds }$

$\dfrac{37\times 46}{100}=\dfrac{34x}{100}+\dfrac{48y}{100}\\\\ 34x+48y=1702$ $-(ii)$

Subtracting 34 times first equation from second equation,

$34x+48y-34x-34y=1702-34\times46\\14y=138\\y=9.857\text{ }pounds \\x=36.143\text{ }pounds$

∴ 36.14 pounds of first alloy and 9.86 pounds of second alloy were used.

5 0

3 years ago