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Schach [20]
3 years ago
12

4 over 5-7 over 15 =

Mathematics
2 answers:
BabaBlast [244]3 years ago
8 0

Answer:

The answer is 1/3.

Tamiku [17]3 years ago
5 0

Answer:

0.33333333333

Step-by-step explanation:

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Yes I can help you with rational expressions and eplquations!
4 0
3 years ago
The larger triangle is the image of the smaller triangle after a dilation. The center of the dilation is (−2,−2). Two triangles
Usimov [2.4K]

Answer:

The scale factor is -1/4.

Step-by-step explanation:

5 0
3 years ago
A) 79.2<br> C) 11.9<br> B) 8.9<br> D) 11.6
vova2212 [387]

Answer:

789

Step-by-step explanation:

8 0
3 years ago
How many times longer is the wavelength of a sound wave with a frequency of 20 waves per second than the wavelength of a sound w
sergejj [24]

The sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec

<h3>Calculating wavelength </h3>

From the question, we are to determine how many times longer is the first sound wave compared to the second sound water

Using the formula,

v = fλ

∴ λ = v/f

Where v is the velocity

f is the frequency

and λ is the wavelength

For the first wave

f = 20 waves/sec

Then,

λ₁ = v/20

For the second wave

f = 16,000 waves/sec

λ₂ = v/16000

Then,

The factor by which the first sound wave is longer than the second sound wave is

λ₁/ λ₂ = (v/20) ÷( v/16000)

= (v/20) × 16000/v)

= 16000/20

= 800

Hence, the sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec

Learn more on Calculating wavelength here: brainly.com/question/16396485

#SPJ1

6 0
2 years ago
Pls solve the simultaneous equation in the attachment.
siniylev [52]

Answer:

Part a) The solution is the ordered pair (6,10)

Part b) The solutions are the ordered pairs (7,3) and (15,1.4)

Step-by-step explanation:

Part a) we have

\frac{x}{2}-\frac{y}{5}=1 ----> equation A

y-\frac{x}{3}=8 ----> equation B

Multiply equation A by 10 both sides to remove the fractions

5x-2y=10 ----> equation C

isolate the variable y in equation B

y=\frac{x}{3}+8 ----> equation D

we have the system of equations

5x-2y=10 ----> equation C

y=\frac{x}{3}+8 ----> equation D

Solve the system by substitution

substitute equation D in equation C

5x-2(\frac{x}{3}+8)=10

solve for x

5x-\frac{2x}{3}-16=10

Multiply by 3 both sides

15x-2x-48=30

15x-2x=48+30

Combine like terms

13x=78

x=6

<em>Find the value of y</em>

y=\frac{x}{3}+8

y=\frac{6}{3}+8

y=10

The solution is the ordered pair (6,10)

Part b) we have

xy=21 ---> equation A

x+5y=22 ----> equation B

isolate the variable x in the equation B

x=22-5y ----> equation C

substitute equation C in equation A

(22-5y)y=21

solve for y

22y-5y^2=21

5y^2-22y+21=0

Solve the quadratic equation by graphing

The solutions are y=1.4, y=3

see the attached figure

<em>Find the values of x</em>

For y=1.4

x=22-5(1.4)=15

For y=3

x=22-5(3)=7

therefore

The solutions are the ordered pairs (7,3) and (15,1.4)

3 0
4 years ago
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