2/5(z+1)=y solve for z

Answers

Answer 1
Answer:

Solving for z, we get → z =  (5y -2)/(2).

We have the following equation -

$y = (2)/(5) (z+1)

We have solve for z.

Solve for x :  ax + \pi = \beta x

Solving this we will get -

βx - ax = \pi

x(β - a) = \pi

x =  (\pi )/(\beta -a)

According to the question, we have -

$y = (2)/(5) (z+1)

⇒ 5y = 2(z + 1)

(5y)/(2) = z + 1

⇒ z = (5y)/(2) - 1

z =  (5y -2)/(2)

Hence, after solving for z, we get → z =  (5y -2)/(2)

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Answer 2
Answer: (2)/(5) (z+1)= y \n \n  (2(z+1))/(5) = y \n \n 2(z + 1) = y * 5 \n \n 2(z + 1) = 5y \n \n z + 1 =  (5y)/(2) \n \n z =  (5y)/(2) -1 \n \n Answer: \fbox {z = 5y/2 - 1}

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Determine whether the vectors u and v are parallel, orthogonal, or neither. u = <7, 2>, v = <21, 6>

Answers

Vectors are orthogonal if their dot product is equal to zero:
u · v = 7· 21 + 2 · 6 = 147 + 12 = 159 ≠ 0
Vectors are parallel if the angle between them is 0°, or cos ( u, v ) = 1
cos  ( u, v ) = ( u · v ) / ( | u | · | v | ) =
= 159 / (√(7² + 2²) · √(21² + 6²) ) =
= 159 / ( √ (49 + 4 ) · √ ( 441 + 36 ) ) = 
= 159 / (√53 · √477 ) = 159 / 159 = 1
Answer : The vectors u and v are parallel. 

Answer:

Parallel

Step-by-step explanation:

orthogonal: dot product of u times v= 0 degrees

utimesv= 7x21+2x6=147+12=159

Parallel: cos (u,v)=1

cos (u,v)=(u·v)/ IuI·IvI

cos (u,v)= 159/ (sqrt 7^2+2^2)·(sqrt 21^2 +6^2)

159/(sqrt 49+4) ·(sqrt 441+36)

159/(sqrt 53)·(sqrt 477)=

159/159

=1; therefore it is parallel

One catalog offers a jogging suit in two colors, gray and black. It comes in sizes S, M, L, XL and XXL. How many possible jogging suits can be ordered? PLEASE HELP NO LINKS!!!

Answers

Answer:

5..

Step-by-step explanation:

How do you solve...0=-5t^2+20t+1

Answers

there were 2 solutions that i came up with. Here is the first one. I rearranged the  eqaution by subtracting what is to the right of thr equal sign. Multiply the coefficeint of the first term by the constant 5 x (-1)= -5. Then you would find 2 factors of -5 whose sum equals the coefficient of the middle term which is 20. -5+1= -4 and  -1+5=4.  the eqaution then comes to 5t^2-20t-1=0. you would solve  5t^2-20t-1=0. you would divide both sides of the equal sign by 5. t^-4t-1/5=0. then t^2-4t=1/5. add 4 to both sides of the equation.  so we get 21/5+ 4 + 4t -t^2=21/5. it then comes out to be t^2-4t+t=t-2^2. according to the law of transitivity it is t-2^2=21/5.

 t =(20+√420)/10=2+1/5√ 105 = 4.049 

It rained 44 inch on Mondayand 54 inch on Tuesday.
How much did it rain on
both days?

30 points

Answers

98 inches, we added the inches.

Answer:

98 inches

Step-by-step explanation:

54inches + 44inches =

5 + 4 = 9  

4 + 4 = 8

So... the answer is 98 inches?  

I hope this helps!!

What's 0.003 1/10 of

Answers

Simple. 
Just multiply .003 by 10.

0.003 x 10 = 0.00030, or 0.0003.

0.0003 is 1/10 of 0.003

A rectangular paperboard measuring 26 in long and 20 in
wide has a semicircle cut out of it.

Find the area of the paperboard that remains. Use the value 3.14
for π, and do not round your answer. Be sure to include the correct unit in your answer.

Answers

Answer:

The remaining area of the paperboard is 363 square inches.

Step-by-step explanation:

The length of the rectangular board = 26 inches

The width of the rectangular board = 20 inches.

Now, the area of the board = 26*20=520 square inches

Now, a semicircle has been cut out of it.

Let is assume that the diameter of the semicircle be 20 inches.

So, radius will be = 10 inches

Area of semicircle is = (1)/(2) \pi r{2}

So, we get : Area =  (1)/(2)*3.14*10*10

= 157 square inches.

Now after cutting the semi circle, the area remained = 520-157=363 square inches.

Therefore, the remaining area of the paperboard is 363 square inches.

Answer:

See below.

Step-by-step explanation:

Find the area of the rectangle using A = l*w = 26*20 = 520.

Now find the area of the semi circle using A = 1/2 πr². Since the instructions do not say the diameter of the semi circle assume it is 20 or 26.

If it is 20: r = 10 and the area is A = 1/2 π10²=50π = 157.

If it is 26: r = 13 and the area is A = 1/2π13² = (84.5)π = 265.33.

The area of the paperboard that remains will be the area of the rectangle minus the area of the semi circle.

If the diameter of the semi circle is 20 then 520 - 157 = 363.

If the diameter of the semi circle is 26 then 520 - 254.67.