All categories

3 years ago

Solve 400 = 8b + y for y

1 answer:

7 0

400 = 8b + y (subtract 8b from each side)
-8b -8b 

400 - 8b = y

The answer is y = -8b + 400

i isolated y, which means i left it alone on one side of the equation

You might be interested in

Solve the linear equation below for x. * 3x + 4 = 8x – 9 A. 2.6 B. 1 C. 1.09 D. 2

Juli2301 [7.4K]

This answer is A, first you subtract 3x from both sides simplify 8x-9-3x to 5x-9, add 9 to both sides, simplify 4+9 to 13, divided both sides by 5 and you’ll get 2.6

6 0

3 years ago

What is the answer to 8s-11s+6

slega [8]

Answer:

-3s + 6

Step-by-step explanation:

Combine like terms. Like terms are terms with the same amount of the same variables. In this case:

8s - 11s + 6

(8s - 11s) + 6

-3s + 6

-3s + 6 is your answer.

7 0

3 years ago

Read 2 more answers

Desmond really like Legos. Last Saturday Desmond counted his Legos and realized that

Burka [1]

Answer: 14 red, 7 green, 44 blue

Step-by-step explanation:

First, use the letter r as a variable to represent the number of red Legos. The number of green Legos (g) is 7 less than the number of red Legos, or g = r-7. The number of blue Legos (b) is 2 more than 3 times the number of red Legos, or b = 3r+2. The total number of Legos is the number of red + green + blue Legos, which can be represented as 65 = r+g+b.

Substitute the equations for g and b in. This should give you a final equation of 65 = r+(r-7)+(3r+2). To solve for the number of red Legos, first add up all of the terms to get 65 = 5r-5. Now add 5 to each side (70 = 5r). Finally, divide each side by 5 (r = 14).

To find the number of green Legos, substitute the number of red Legos (14) into the equation for the green Legos (g = r-7). This should get you the equation g = 14-7 which simplifies to g = 7.

To find the number of blue Legos, substitute the number of red Legos (14) into the equation for the blue Legos (b = 3r+2). This gives you the equation b = (3*14)+2. First, multiply 3 and 14 to get b = 42+2. Finally, add them together to get b = 44.

6 0

3 years ago

Can someone help on a construction of a tangent to a circle please, need pictures if possible

Anit [1.1K]

Wassup

Steps:

1. Draw a line connecting the point to the center of the rectangle

2. Then construct a perpendicular bisector to the line drawn in step 1

3. Place the compass on the midpoint of the line, adjust its length to reach the end point, and draw an arc across the circle

4. Where the arc crosses the circle will be the tangent points. So connect the endpoint to the tangent point

Hope the picture helps you too

5 0

3 years ago

Find the length of the curve y = integral from 1 to x of sqrt(t^3-1)

Arlecino [84]

$y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt$

By the fundamental theorem of calculus,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}$

Now the arc length over an arbitrary interval $(a,b)$ is

$\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx$

But before we compute the integral, first we need to make sure the integrand exists over it. $x^{3/2}$ is undefined if $x$ , so we assume $a\ge0$ and for convenience that $a$ . Then

$\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)$

6 0

3 years ago