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Andrej [43]
3 years ago
5

Cyrus solved a quadratic equation. His work is shown below.

Mathematics
2 answers:
ASHA 777 [7]3 years ago
7 0

Answer:

sorry the text is all messsed up so i cant answer the questionstep #

Brrunno [24]3 years ago
3 0

Answer:

Step 2

Step-by-step explanation:

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PLEASE HELP SOLVE FOR ‘X’!!!
antoniya [11.8K]

Answer:

x= 10

Step-by-step explanation:

1/5 x -2/3 = 4/3

Add 2/3 to each side

1/5x -2/3 +2/3 = 4/3 +2/3

1/5x = 6/3

1/5x = 2

Multiply each side by 5

1/5x * 5 = 2*5

x = 10

3 0
3 years ago
F(x)=bx^2+32 For the function f defined above, b is a constant and f(2)=40. What is the value of f(-2)?
zhuklara [117]

Answer:

f(-2) = 0

Step-by-step explanation:

Given that:

f(x) = bx^2 + 32

f(2) = b(2)^2 + 32

f(2) = 4b + 32

f(2) = 4b = -32

f(2) = b = -32/4

f(2) = b = -8

Thus;

f(-2) = -8(-2)^2 + 32

f(-2) = -8(4) + 32

f(-2) = -32 + 32

f(-2) = 0

6 0
3 years ago
Read 2 more answers
Identify the diameter of the circular base created by folding the figure into a right cone. HELP ASAP PLEASE!!
Akimi4 [234]

let's notice something, we have a circle with a radius of 12 and one 90° sector is cut off, so only three 90° sectors of the circle are left shaded, so namely the cone will be using 3/4 of that circle.

think of it as, this shaded area is some piece of paper, and you need to pull it upwards and have the cutoff edges meet, and when that happens, you'll end up with a cone-shaped paper cup, and pour in some punch.

now, once we have pulled up the center of the circle to make our paper cup, there will be a circular base, its diameter not going to be 24, it'll be less, but whatever that base is, we know that is going to have the same circumference as those in the shaded area.  Well, what is the circumference of that shaded area?

\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi

well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?

\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill

3 0
3 years ago
What is the axis of symmetry of the
S_A_V [24]

Answer:

A

Step-by-step explanation:

i think so..sorry if im wrong

8 0
3 years ago
Read 2 more answers
Prove that the triangle EDF is isosceles. Give reasons for your answer.
Gekata [30.6K]

Answer:

\triangle EDF is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

\text{External }\angle DFG = 90 +\dfrac{y}{2}

Let us try to find out \angle E and \angle DFE. After that we will compare them.

<u>Finding </u>\angle DFE<u>:</u>

Side EG is a straight line so \angle GFE = 180

\angle GFE is sum of internal \angle DFE and external \angle DFG

\angle GFE = 180 = \angle DFE  + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)

<u>Finding </u>\angle E<u>:</u>

<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external \angle DFG = \angle E + \angle EDF

\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2}  -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)

Comparing equations (1) and (2):

It can be clearly seen that:

\angle DFE = \angle E =90-\dfrac{y}{2}

The two angles of \triangle EDF are equal hence \triangle EDF is isosceles.

8 0
3 years ago
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