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2 years ago

Sarah solved a quadratic equation. Her work is shown below, with Step 1 missing.

Mathematics

2 answers:

malfutka [58]2 years ago

7 0

she Could have wriiten 2(9+1)^2+6+134 did that help!!!!!!!!!??????????

to check maybe u should try it jsut caculate it :D

KengaRu [80]2 years ago

3 0

Answer:

step 1 - subtract 6 from both sides

Step-by-step explanation:

Given

2(x + 1)² + 6 = 134

Step 1 - subtract 6 from both sides

2(x + 1)² = 128

Step 2 - divide both sides by 2

(x + 1)² = 64

Step 3 - take the square root of both sides

x + 1 = ± $\sqrt{64}$ = ± 8

Step 4 - subtract 1 from both side

x = - 1 ± 8

Step 5 - solve for x

x = - 1 - 8 = - 9 or x = - 1 + 8 = 7

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How many times greater is 7 x 10^6 than 4.5 x 10^-1

kipiarov [429]

Let us assume that : 7 × 10⁶ is P times Greater than 4.5 × 10⁻¹

It means when we Multiply P with 4.5 × 10⁻¹, We should get 7 × 10⁶

⇒ (P × 4.5 × 10⁻¹) = 7 × 10⁶

$\mathsf{\implies P = (\dfrac{7 \times 10^6}{4.5 \times 10^-^1})}$

$\mathsf{\implies P = 1.56 \times 10^7}$

⇒ 7 × 10⁶ is 1.56 × 10⁷ times Greater than 4.5 × 10⁻¹

4 0

2 years ago

Type the correct answer in each box.

Mariulka [41]

Answer:

$(x-5)^2+(y+4)^2=100$

Step-by-step explanation:

step 1

Find the radius of the circle

we know that

The distance between the center and any point that lie on the circle is equal to the radius

we have the points

(5,-4) and (-3,2)

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute the values

$r=\sqrt{(2+4)^{2}+(-3-5)^{2}}$

$r=\sqrt{(6)^{2}+(-8)^{2}}$

$r=\sqrt{100}\ units$

$r=10\ units$

step 2

Find the equation of the circle

we know that

The equation of a circle in standard form is equal to

$(x-h)^2+(y-k)^2=r^2$

where

(h,k) is the center

r is the radius

we have

$(h,k)=(5,-4)\\r=10\ units$

substitute

$(x-5)^2+(y+4)^2=10^2$

$(x-5)^2+(y+4)^2=100$

6 0

3 years ago

Find the domain of the function

Svetradugi [14.3K]

Answer:

<em>The domain of f is (-∞,4)</em>

Step-by-step explanation:

<u>Domain of a Function</u>

The domain of a function f is the set of all the values that the input variable can take so the function exists.

We are given the function

$f(x)=\frac{1}{\sqrt{4-x} }$

It's a rational function which denominator cannot be 0. In the denominator, there is a square root whose radicand cannot be negative, that is, 4-x must be positive or zero, but the previous restriction takes out 0 from the domain, thus:

4 - x > 0

Subtracting 4:

- x > -4

Multiplying by -1 and swapping the inequality sign:

x < 4

Thus the domain of f is (-∞,4)

7 0

2 years ago

you took out a loan of $1,200 with the principal pay off of $200 a month each month you paid 200 plus the interest at a rate of

Bad White [126]

I think the answer you are looking for is $2,592

5 0

2 years ago

Need help<br> Question in the screenshot below<br> NO spam<br> Please show your work

Alja [10]

<h3>Answer: x = 6</h3>

======================================================

Work Shown:

$\log_{4}(x+10)+\log_{4}(x-2)=\log_{4}(64)\\\\\log_{4}\left((x+10)(x-2)\right)=\log_{4}(64)\\\\(x+10)(x-2)=64\\\\x^2-2x+10x-20=64\\\\x^2-2x+10x-20-64=0\\\\$

$x^2+8x-84=0\\\\(x+14)(x-6)=0\\\\x+14=0 \ \text{ or } \ x-6=0\\\\x=-14 \ \text{ or } \ x=6\\\\$

Those are the possible solutions, but plugging x = -14 back into the original equation will lead to an error. So we rule x = -14 out

x = 6 works as a solution however

3 0

2 years ago