How do you solve question 27 and 28

Mathematics

2 answers:

Brrunno [24]3 years ago

6 0

Answer:#1 C.30 #2 40

Step-by-step explanation:

3x+15=7x-5

3x+15+5=7x-5+5

3x+20=7x

3x-3x+20=7x-3x

20=4x

20/4=4/4x

5=x so if it is an equilateral all sides are equal so 3(5)+15=30 so the answer should be 30

So the triangle in #2 is an isoceles so two of the angles are equal so the equation is 2(2x-24)+3x+4=180

4x-48+3x+4=180

7x-44=180

7x=224

x=32

2(32)-24

64-24=40

Nataly_w [17]3 years ago

5 0

You solve it like the person up there said people are really smart these days lol

You might be interested in

The function f(x) varies inversely with x and f(x) 0.2 when x= 0.4.

zhannawk [14.2K]

I hope this helps you

7 0

3 years ago

Read 2 more answers

Rewrite the function by completing the square. 4x^2-4x+1=0

andrey2020 [161]

Do in completing the square method.

${4x}^{2} - 4x + 1 = 0 \\ 4x^{2}-4x=-1 \\ \frac{4x^{2}-4x}{4}=\frac{-1}{4} \\ x^{2}+\frac{-4}{4}x=\frac{-1}{4} \\ x^{2}-x=\frac{-1}{4} \\ x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{1}{4}+\left(-\frac{1}{2}\right)^{2} \\ x^{2}-x+\frac{1}{4}=\frac{-1+1}{4} \\ x^{2}-x+\frac{1}{4}=0 \\ \left(x-\frac{1}{2}\right)^{2}=0 \\ \sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{0}$

Now simplify.

$x-\frac{1}{2}=0 \\ x-\frac{1}{2}=0 \\ \\ \rightarrow \: x=\frac{1}{2} \\ \rightarrow \: x=\frac{1}{2}$

So, answer is..

$\boxed{x=\frac{1}{2} }$

Now rewrite the equation.

$4 x ^ { 2 } - 4 x + 1 = 0 \\ \boxed{4\times \left(\frac{1}{4}\right)-4\times \left(\frac{1}{2}\right)+1=0 }$

Verification :-

$4 x ^ { 2 } - 4 x + 1 = 0 \\ 4\times \left(\frac{1}{4}\right)-4\times \left(\frac{1}{2}\right)+1=0 \\ 1-4\times \left(\frac{1}{2}\right)+1=0 \\ 1-\frac{4}{2}+1=0 \\ 1-2+1=0 \\ -1+1=0 \\ \underline{ 0 = 0} \: \\ \dashrightarrow \text{true}$