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3 years ago

Guys how do you work what x and y is good you please explain

Mathematics

1 answer:

Mkey [24]3 years ago

4 0

Answer:

x=10 y=2

Step-by-step explanation:

4x-2y=36 x2

5x-4y=42

8x-4y=72

5x-4y=42

3x-0=30

30÷3=10

x=10

(4x10)-2y=36

40-36=4

4÷2=2

y=2

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Two supplementary angles differ by 34. find the angles..

antiseptic1488 [7]

Answer:

107°

73°

Step-by-step explanation:

Let the two supplementary angles be x° and y°.

x + y = 180....(1)

Since, supplementary angles differ by 34.

Therefore,

x - y = 34....(2)

Adding equations (1) & (2)

x + y = 180

x - y = 34

_________

2x = 214

x = 214/2

x = 107°

Plug x = 107 in equation (1)

107 + y = 180

y = 180- 107

y = 73°

3 0

3 years ago

22) Find the measure of angle A to the nearest degree. Find the length of side b, to tenths.

mr Goodwill [35]

Answer:

m∠A = 31

b = 13.3

Step-by-step explanation:

5 0

3 years ago

The simplified expression

Romashka-Z-Leto [24]

Answer:

$5x^2 y^2$

Step-by-step explanation:

We need to use the properties shown below to solve this:

1. $\sqrt[n]{x^a} =x^{\frac{a}{n}}$

2. $\sqrt{x}\sqrt{x} =x$

3. $\sqrt{x} \sqrt{y}=\sqrt{x*y}$

Area of a triangle is given by 1/2 * base * height, so we do that and simplify:

$A=\frac{1}{2}(\sqrt{5x^3} )(2\sqrt{5xy^4} )\\A=\frac{1}{2}(5x^3)^{\frac{1}{2}}*2*(5xy^4)^{\frac{1}{2}}\\A=\sqrt{5}x^{\frac{3}{2}}*\sqrt{5}\sqrt{x} } y^2\\A=\sqrt{5} \sqrt{5}x^{\frac{3}{2}} x^{\frac{1}{2}}y^2\\A=5*x^2y^2\\A=5x^2 y^2$

6 0

3 years ago

A group of student organized a car wash to raise money for a local charity.the student charged $5 for each car they washed. In 3

Alex Ar [27]

$5 each car 3 hours = 12 cars ÷3 ÷3 1 hours = 4 cars ×8 ×8 8 hours = 32 cars 32 cars × $5 per car = $160 They earned $160 in 8 hours, from washing 32 cars.

7 0

3 years ago

PLEASE HELPP I AM FAILING AND I NEED THIS ASAP!!! TYSM

Lana71 [14]

Answer:

See below for answers

Step-by-step explanation:

Do you mean $f(x)=2sin(x-1)$ or $f(x)=2sin(x) -1$ ?

Use the formula $f(x)=asin(bx+c)+d$ where $|a|$ is the amplitude, $-\frac{c}{b}$ is the phase shift, $d$ is the vertical shift, and the period is $\frac{2\pi}{|b|}$ .

Assuming your first case is true, the amplitude would be $|a|=|2|=2$ , your period is $\frac{2\pi}{|1|}=2\pi$ , and your midline/vertical shift is $d=0$ .

Assuming your second case is true, everything but the vertical shift stays the same. It would be $d=-1$ in this case.

5 0

3 years ago