1. Jared has portrait in shape of a square. One side length, in inches, is the sum of 7 and L

Mathematics

1 answer:

Fed [463]3 years ago

5 0

Answer:

7 + L + 3 = L + 10

Step-by-step explanation:

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Select the letter of the correct answer.

sveta [45]

Would it be B which is 51%
as you do 129 devided by 3.3 = 39.1
180-39.1= 141
141 is just over half

4 0

3 years ago

Read 2 more answers

Solve the proportion

mario62 [17]

Answer:

$\boxed{\sf x = 5}$

Step-by-step explanation:

$\sf Solve \: for \: x \: over \: the \: real \: numbers: \\ \sf \implies \frac{2}{x - 3} = \frac{5}{x} \\ \\ \sf Take \: the \: reciprocal \: of \: both \: sides: \\ \sf \implies \frac{x - 3}{2} = \frac{x}{5} \\ \\ \sf Expand \: out \: terms \: of \: the \: left \: hand \: side: \\ \\ \sf \implies \frac{x}{2} - \frac{3}{2} = \frac{x}{5} \\ \\ \sf Subtract \: \frac{x}{5} - \frac{3}{2} \: from \: both \: sides: \\ \sf \implies \frac{x}{2} - \frac{3}{2} - ( \frac{x}{5} - \frac{3}{2} ) = \frac{x}{5} - ( \frac{x}{5} - \frac{3}{2} ) \\ \\ \sf \implies \frac{x}{2} - \frac{3}{2} - \frac{x}{5} + \frac{3}{2} = \frac{x}{5} - \frac{x}{5} + \frac{3}{2} \\ \\ \sf \frac{x}{5} - \frac{x}{5} = 0 : \\ \sf \implies \frac{x}{2} - \frac{x}{5} - \frac{3}{2} + \frac{3}{2} = \frac{3}{2} \\ \\ \sf \frac{3}{2} - \frac{3}{2} = 0: \\ \sf \implies \frac{x}{2} - \frac{x}{5} = \frac{3}{2} \\ \\ \sf \frac{x}{2} - \frac{x}{5} = \frac{5x - 2x}{10} = \frac{3x}{10} : \\ \sf \implies \frac{3x}{10} = \frac{3}{2} \\ \\ \sf Multiply \: both \: sides \: by \: \frac{10}{3} : \\ \sf \implies \frac{3x}{10} \times \frac{10}{3} = \frac{3}{2 } \times \frac{10}{3} \\ \\ \sf \frac{3x}{10} \times \frac{10}{3} = \cancel{\frac{3}{10} } \times( x) \times \cancel{ \frac{10}{3} } = x : \\ \sf \implies x = \frac{3}{2} \times \frac{10}{3} \\ \\ \sf \frac{3}{2} \times \frac{10}{3} = \cancel{ \frac{3}{2} } \times \cancel{ \frac{3}{2} } \times 5 : \\ \sf \implies x = 5$

8 0

3 years ago

How do I solve for x

Alchen [17]

Greetings!

To find the length of any side of a right triangle, you can use the Pythagorean Thereom. It states that the squares of two sides are equal to the square of the hypotenuse:
$a^2+b^2=c^2$

Input the information from the diagram into the formula:
$(x)^2+(x+7)^2=(13)^2$

Expand each term:
$(x)^2+(x+7)^2=(13)^2$

$x^2+((x+7)(x+7))=169$

$x^2+(x(x+7)+7(x+7))=169$

$x^2+x^2+7x+7x+49=169$

Combine like terms:
$2x^2+14x+49=169$

Add -169 to both sides:
$(2x^2+14x+49)+(-169)=(169)+(-169)$

$2x^2+14x-120=0$

Factor out the Common Term (2):
$2(x^2+7x-60)=0$

Factor the Complex Trinomial:
$2(x^2-5x+12x-60)=0$

$2(x(x-5)+12(x-5))=0$

$2(x-5)(x+12)=0$

Set Factors to equal 0:
$x-5=0$

$x=5$

or

$x+12=0$

$x=-12$

However, since we are solving for the side length, the only possible answer is 5 (a shape can't have a side with a negative length.)

The Solution Is:
$\boxed{x=5}$

I hope this helped!
-Benjamin

6 0

3 years ago

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Murljashka [212]

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