How to make my answer for 0.70 a fraction

Answer:

Step-by-step explanation:

Slope of line through (0,-1) and (1,2) = (-1 - 2)/(0 - 1) = 3

Point-slope equation for line of slope 3 that passes through (0,-1):

y+1 = 3(x-0)

When x = 2:

y+1 = 3(2-0)

y = 3·2 - 1 = 5

When x = 3:

y+1 = 3(3-0)

y = 3·3-1 = 8

Step-by-step explanation:

a) let number=x

four times a number=4x

Condition:

4x+6=50

4x=50-6

4x=44

x=44/4

x=11

b) Condition:

x+9×-2=-8

x-18=-8

x=-8+18

x=10

Note:if you need to ask any question please let me know.

Find the sale price by multiplying the original price by 70% then add the two prices together to get the total.

799 x 0.70 = 559.30

549 x 0.70 = 384.30

Total: 559.30 + 384.30 = $943.60

[tex]\\ \Large\sf\longmapsto h(t)[/tex]

[tex]\\ \Large\sf\longmapsto 2+8t-3t^2+\dfrac{1}{5}t^3[/tex]

[tex]\\ \Large\sf\longmapsto 2+8(3)-3(3)^2+\dfrac{1}{5}(3)^3[/tex]

[tex]\\ \Large\sf\longmapsto 2+24-3(9)+\dfrac{27}{5}[/tex]

[tex]\\ \Large\sf\longmapsto 26-27+5.4[/tex]

[tex]\\ \Large\sf\longmapsto -2+5.4[/tex]

[tex]\\ \Large\sf\longmapsto h(t)=3.4m[/tex]

Answer:

ask in English then I can help u

Answer:

GHC22.5

Step-by-step explanation:

90/3=30

30=0.75

30×0.75

=22.5

Answer:

incomplete question

Step-by-step explanation:

that is what is wrong with your question

Answer:

r = 4.3%

Step-by-step explanation:

6810= 6000(x)^3

6810/6000=  (x)^3

x = 1.043114431

r = 043114431

Answer:

3. 6a+60

5. 25+5w

7. 90-10t

11. 4.5-12c

13. f-2

15. 12z+1.5

Step-by-step explanation:

3.

6(a+10)

Multiply 6 by both factors in the parentheses, in this case, a and 10.

6*a = 6a

6*10 = 60

6(a+10) = 6a + 60

I only put the step- by- step explanation for #3, but you should be able to figure the rest out with that.

The length of the confidence interval is the margin of error, which is the ratio of the standard deviation and the square root of sample size. Hence, to reduce the length of confidence interval by half, Quadruple the sample size.

Recall :

Evaluating an hypothetical scenario :

Let standard deviation, σ = 2

Sample size = 50

Margin of Error = 2/√50 = 0.554

Using Quadruple of the sample size : (50 × 4) = 200 samples

(0.227 ÷ 0.554) = 0.5

Therefore, increasing the sample size, reduces the margin of error. Hence, using quadruple the sample size, will reduce the margin of error by half.

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Answer:

[tex]P(C'\ and\ H') =0. 2178[/tex]

Step-by-step explanation:

Let

[tex]C \to[/tex] Student with cat

[tex]H \to[/tex] Student has HBO sub

[tex]P(C) = 66\% \\ P(H) = 37\%[/tex]

Required

[tex]P(C'\ and\ H')[/tex]

This is calculated as:

[tex]P(C'\ and\ H') = P(C') * P(H')[/tex]

Using complement rules, we have:

[tex]P(C'\ and\ H') = [1 - P(C)] * [1 - P(H)][/tex]

So, we have:

[tex]P(C'\ and\ H') = [1 - 66\%] * [1 - 37\%][/tex]

[tex]P(C'\ and\ H') = [33\%] * [66\%][/tex]

[tex]P(C'\ and\ H') =0. 2178[/tex]

Answer:

-2

Step-by-step explanation:

Perpendicular lines have slopes that are negative reciprocals

Take the slope of AB = 1/2

-1/(1/2)

-1 * 2/1

-2

The slope of a line perpendicular is -2

Answer:

[tex]A). \ \ \frac{(72\pi + 9\pi )}{4} \ in^2[/tex]

Step-by-step explanation:

Given;

radius of the circle, r = 9 inches

the part of the circle cut out = one-forth of the complete circle

the angle of the sector cut out θ= ¹/₄ x 360 = 90⁰

Area of the complete circle = πr² = π x 9² = 81π in²

Area of the sector cut out = [tex]= \frac{\theta }{360} \pi r^2 = \frac{90}{360} \pi (9^2) = \frac{1}{4} \times 81\pi = \frac{81 \pi}{4} = \frac{(72\pi + 9\pi)}{4} \ in^2[/tex]

Therefore, the only correct option is A. [tex]\frac{(72\pi + 9\pi )}{4} \ in^2[/tex]

Answer:

850,000

Step-by-step explanation:

Answer: 850,000

Concept:

Here, we need to know the order and name of each place value.

Please refer to the attachment below for the specified names.

Solve:

8 = Hundred thousands

4 = Ten thousands

9 = One thousands

7 = Hundreds

0 = Tens

9 = Ones

Since the values before the ten thousands place, which would be the one thousands place, is greater than 5, then we should round up.

Therefore, the rounded value would be [tex]\boxed{850,000}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer:

48 NO seña hfjxsmisns sisbxbd

Step-by-step explanation:

Answer:

30 and 270 respectively

Step-by-step explanation:

Let the speed of plane in still air be x and the speed of wind be y.

ATQ, (x+y)*4=1200 and (x-y)*5=1200. Solving it, we get x=270 and y=30

Answer:

m = [tex]\frac{2K}{v^2}[/tex]

Step-by-step explanation:

Given mass (m ) varies directly as K and inversely as v² then the equation relating them is

m = [tex]\frac{kK}{v^2}[/tex] ← k is the constant of variation

To find k use the condition m = 10 when K = 80 and v = 4 , then

10 = [tex]\frac{80k}{4^2}[/tex] = [tex]\frac{80k}{16}[/tex] ( multiply both sides by 16 to clear the fraction )

160 = 80k ( divide both sides by 80 )

2 = k

m = [tex]\frac{2K}{v^2}[/tex] ← equation of variation

The expression of mass as a function of kinetic energy and velocity is m = 2K/V².

Expressions are defined as mathematical statements that have a minimum of two terms containing variables or numbers.

We have been given that the mass varies directly as the Kinetic energy (K) and inversely as the square of the velocity (V).

m ∝ K/V²

m = cK/V²

Here c is the constant of variation,

If the kinetic energy is 80 Joules and the velocity is 4 meters per second, then the mass is 10 kilograms.

We have to determine the value of c

Here m = 10 , K = 80 and V = 4 , then

Substitute the values in m =  cK/V²

10 = c(80)/4²

10 = c(80)/16

10 = 5c

c = 10/5

c = 2

Substitute the value of c = 2 in the equation of variation,

⇒ m = 2K/V²

Hence, the expression of mass as a function of kinetic energy and velocity is m = 2K/V².

Learn more about the expressions here:

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Answer:

48degrees

Step-by-step explanation:

From the circle geometry shown, traingle BDC is an isosceles triangle which shows means that their base angels are the same. Hence;

<B = <C

<CBD + <BCD + <D = 180

<BCD + <BCD + <D =180

2<BCD + <BDC = 180

Get <BCD;

<BCD+ <ECB = 90

<BCD + 48 = 90

<BCD = 90 - 48

<BCD = 42degrees

Get <BDC

2<BCD + <BDC = 180

2(42)+ <BDC = 180

84 + <BDC = 180

<BDC = 180 - 84

<BDC = 96

Since angle at the centre is twice that at the circumference, then;

<BAC = 1/2(<BDC )

<BAC = 96/2

<BAC = 48degrees

Answer:

$.08 per mile

Step-by-step explanation:

$2.25          gallon

------------ * ---------------

gallon          28 miles

$2.25

-------------

28 miles

$.080357143 per mile

Rounding to the nearest cent

$.08 per mile

9514 1404 393

Answer:

  (a)  {(5, 2), (-4, 2), (3, 6), (0, 4), (-1, 2)}

Step-by-step explanation:

The only relation with no repeated x-values is the first one. The first relation is a function.

Answer:

108 meters squared or m^2

Step-by-step explanation:

* means multiply

15 is probably hypotenuse because its the longest

12 and 9 are probably base and height

area = base * height

area = 12 * 9

area = 108

Answer:

54m^2

Step-by-step explanation:

Answer:

[tex]\cos(11x)[/tex]

Step-by-step explanation:

Given

[tex]\cos 5x\ \cos 6x- \sin\ 5x \sin 6x[/tex]

Required

Express as a single function

In trigonometry, we have:

[tex]\cos(A + B) = \cos A\cos B - \sin A \sin B[/tex]

By comparison, we have

[tex]\cos(5x + 6x) = \cos 5x\cos 6x - \sin 5x \sin 6x[/tex]

[tex]\cos(11x) = \cos 5x\cos 6x - \sin 5x \sin 6x[/tex]

9514 1404 393

Answer:

  4.8 years

Step-by-step explanation:

Solving the compound interest formula for the number of years gives ...

  t = log(A/P)/(n·log(1 +r/n))

where principal P invested at rate r compounded n times per year produces value A after t years.

  t = log(24805/22000)/(365·log(1 +0.025/365)) ≈ 4.800

The loan was for 4.8 years.

z = 3i / (-1 - i )

z = 3i / (-1 - i ) × (-1 + i ) / (-1 + i )

z = (3i × (-1 + i )) / ((-1)² - i ²)

z = (-3i + 3i ²) / ((-1)² - i ²)

z = (-3 - 3i ) / (1 - (-1))

z = (-3 - 3i ) / 2

Note that this number lies in the third quadrant of the complex plane, where both Re(z) and Im(z) are negative. But arctan only returns angles between -π/2 and π/2. So we have

arg(z) = arctan((-3/2)/(-3/2)) - π

arg(z) = arctan(1) - π

arg(z) = π/4 - π

arg(z) = -3π/4

where I'm taking arg(z) to have a range of -π < arg(z) ≤ π.

Answer: [tex]10,000\ lb.ft[/tex]

Step-by-step explanation:

Given

Initial weight of the bucket is [tex]145\ lb[/tex]

It is lifted at constant rate and rate of sand escaping is [tex]0.5\ lb/ft[/tex]

At any height weight of the sand is [tex]w(h)=145-0.5h[/tex]

Work done is given by the product of applied force and displacement or the area under weight-displacement graph

from the figure area is given by

[tex]\Rightarrow W=\int_{0}^{80}\left ( 145-0.5h \right )dh\\\\\Rightarrow W=\left | 145h-\dfrac{0.5h^2}{2} \right |_0^{80}\\\\\Rightarrow W=\left [ 145\times 80-\dfrac{0.5(80))^2}{2} \right ]-0\\\\\Rightarrow W=11,600-1600\\\\\Rightarrow W=10,000\ lb.ft[/tex]

Answer:

40141

Step-by-step explanation: