An isosceles trapezoid has base angles of 45° and bases of lengths 8 and 14. The area of the trapezoid is

How do you solve the indicated variable? 6=mx+b for x

UkoKoshka [18]

$6 = mx + b \\ 6 - b = mx \\ \frac{6 - b}{m} = x$

6 0

3 years ago

Read 2 more answers

Select the correct answer

tatuchka [14]

Answer:

A

Step-by-step explanation:

A real interest rate is an interest rate that has been adjusted to remove the effects of inflation to reflect the real cost of funds to the borrower and the real yield to the lender or to an investor. A nominal interest rate refers to the interest rate before taking inflation into account.

7 0

3 years ago

Read 2 more answers

Simplify. 2^5−4⋅3^3 7+(1+100√)

frez [133]

$\\ \sf\bull\longmapsto 2^5-4(3^3)$

$\\ \sf\bull\longmapsto 32-4(27)$

$\\ \sf\bull\longmapsto 32-108$

$\\ \sf\bull\longmapsto -76$

7 0

2 years ago

bianca is building a sandbox that will hold 4 cubic yards of sand. she wants it to be 18 inches deep and 2 yards wide. what will

tamaranim1 [39]

Length should be 2/5yd:

18inch*1yd/36inch=.5yd=> depth

volume= 2(length*width+lenght*depth+depth*width)

let l to be the length.

volume= 2(l*2+2(.5)+.5*2)

volume=5l+2

4=5l+2

2=5l

l=2/5

5 0

3 years ago

Anyone can help me solve this equation using cross multiplying

Natali5045456 [20]

9514 1404 393

Answer:

x = 1 or 5

Step-by-step explanation:

The notion of "cross-multiplying" is the idea that the numerator on the left is multiplied by the denominator on the right, and the numerator on the right is multiplied by the denominator on the left. This looks like ...

$\displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\ \longrightarrow\ (x-1)(3x-1)=(7)(2x-2)$

Then the solution proceeds by eliminating parentheses, and solving the resulting quadratic equation.

$3x^2-4x+1=14x-14\\\\3x^2-18x+15=0\qquad\text{subtract $14x-14$}\\\\x^2-6x+5=0 \qquad\text{divide by 3}\\\\(x-1)(x-5)=0\qquad\text{factor}\\\\x\in\{1,5\}$

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Comment on "cross multiply"

Like a lot of instructions in Algebra courses, the idea of "cross multiply" describes what the result looks like. It doesn't adequately describe how you get there. The one and only rule in solving Algebra problems is "whatever is done to one side of the equation must also be done to the other side of the equation." If you multiply one side by one thing and the other side by a different thing, you are violating this rule.

What looks like "cross multiply" is really "multiply by the product of the denominators and cancel like terms from numerator and denominator." Here's what that looks like with the intermediate steps added.

$\displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\\\\\frac{x-1}{7}\times7(3x-1)=\frac{2x-2}{3x-1}\times7(3x-1)\\\\(x-1)(3x-1)=(2x-2)(7)\qquad\textit{looks like}\text{ cross multiply}$

8 0

2 years ago