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LenaWriter [7]
3 years ago
7

What is the area of this figure? A 28 yd² B 40 yd C52 yd² D64 yd²

Mathematics
2 answers:
Fudgin [204]3 years ago
5 0
The BEST answer for this QUESTION is:
*(D. 64 yd² 
d1i1m1o1n [39]3 years ago
3 0
<span>D64 yd² i hope this helps</span>
You might be interested in
How do you write 3.25 in simplest form
dedylja [7]
3.25 would be 3 1/4 because,

Step 1: You would rewrite the decimal number as a fraction with 1 in the denominator. So, 3.25 = 3.24/1

Step 2: Multiply it by 1 to eliminate 2 decimals places, so multiply the top and bottom by 10^2 = 100
3.25/1 x 100/100 = 325/100
Step 3: Now you find the greatest common factor, also known as GCF, of 325 and 100 if it exits then you reduce the fraction by dividing both numerator and denominator by it. GCF = 25
3.25/100 divided by 25/25 = 13/4
Now you would simplify it to make a proper fraction getting your answer
3 1/4
Does that make sense?
7 0
3 years ago
Read 2 more answers
Write the product using exponents. 1/5 x 1/5 x 1/5 x 1/5 x 1/5
damaskus [11]

Answer: 1/5 with an exponent of 5.

hope this helps :)

7 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\}

_____

<em>Comment on "cross multiply"</em>

Like a lot of instructions in Algebra courses, the idea of "cross multiply" describes <em>what the result looks like</em>. It doesn't adequately describe how you get there. The <em>one and only rule</em> in solving Algebra problems is "<em>whatever is done to one side of the equation must also be done to the other side of the equation</em>." 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 "<em>multiply by the product of the denominators</em> 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
3 years ago
How do you solve an inequality using an absolute value?<br> Example: |x+3|&lt;2
ICE Princess25 [194]

Answer:

x < -1

Step-by-step explanation:

hope that will help you

4 0
3 years ago
3x^2 - 12x ÷ x-4 what is this simplified
Andreyy89
<span>3<span>x^2</span></span>−<span>16 I think that's the answer. Hope that helps!</span>
8 0
3 years ago
Read 2 more answers
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