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3 years ago

14

Find the value of x. 45 60 90 80

1 answer:

IRISSAK [1]3 years ago

4 0

Answer:

Step-by-step explanation:

The formula you want is x°= $\frac{1}{2}$ (major arc - minor arc)

If the minor arc is 90, and all the way around the outside of a circle is 360 degrees, that means that the major arc measures 360 - 90 = 270. Filling in our formula:

x°= $\frac{1}{2}$ (270 - 90)

or

x° = $\frac{1}{2}$ (180)

That gives us that x is a right angle at 90°

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If 6y-8=28, what is the value of y

quester [9]

The answer would be y = 6

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the interest on an investment varies directly as the rate of of interest. if interest is $47 when interst rate is 5% find the in

NNADVOKAT [17]

Answer:

The interest when rate is 5.2% is $48.88

Step-by-step explanation:

if interest (I) varies directly as interest rate (R)

Then

I ∝ R

I=cR

c= constant

if interest is $47 when interest rate is 5%

47 =.05c

c = 940

if interest rate is 5.2% then

I = 940 x .052

=48.88

=$48.88

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3 years ago

Write the decimal 0.36 recurring as a fraction in simplest form.

Grace [21]

Answer:

36/100

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3 years ago

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1. What is the vertex form of the equation?

sergeinik [125]

1.vertex form is $y=a(x-h)^2+k$

to get into vertex form, complete the square

$y=-x^2+12x-4$

step 1: group x terms

$y=(-x^2+12x)-4$

step 2: factor coefient in front of x^2 term

$y=-1(x^2-12x)-4$

prep for next step: take 1/2 of linear (1st dgree term) and square it

$\frac{-12}{2}=-6$ , $(-6)^2=36$

step 3: add positive and negative of that previous number inside the parenthasees

$y=-1(x^2-12x+36-36)-4$

factor perfect squaer trionmial

$y=-1((x-6)^2-36)-4$

take out that -36 by expansion

$y=-1(x-6)^2+36-4$

$y=-1(x-6)^2+32$

2. factor out the 2

$2(x^2+8x+12)$

what 2 numbers multiply to get 12 and add to get 8?

2 and 6

$2(x+2)(x+6)$

3.

for $ax^2+bx+c=0$

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

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a=1, b=-7, c=-6

$x=\frac{-(-7) \pm \sqrt{(-7)^2-4(1)(-6)}}{2(1)}$

$x=\frac{7 \pm \sqrt{73}}{2}$

so $x=\frac{7+\sqrt{73}}{2}$ and $x=\frac{7-\sqrt{73}}{2}$

4. remember that $i=\sqrt{-1}$ so $i^2=-1$

treat $i$ as a variable

$(-2i)(8i)=$

$(-2)(i)(8)(i)=$

$(-2)(8)(i)(i)=$

$(-16)(i^2)=$

$(-16)(-1)=$

[tex}16[/tex]

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Fudgin [204]

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3 years ago