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3 years ago

Please help me I need help right away.

Mathematics

2 answers:

ra1l [238]3 years ago

8 0

Answer:

3 7/16

Step-by-step explanation:

o-na [289]3 years ago

4 0

Answer:

Step-by-step explanation:

improper fraction you take the whole number, multiply it by the denominator and add the numerator. That number goes on top and you keep the denominator the sam.

2 1/2 in an improper fraction is 5/2

1 3/8 in an improper fraction is 11/8

keep change flip

keep the first fraction

change multiplication to division and flip the second fraction

5/2 x 11/8

5/2 divided by 8/11

answer is 3 7/16

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Ellen is decorating a ballroom ceiling with garland. If the rectangular ceiling is 48 feet by 64 feet, how much garland will Ell

balandron [24]

80 feet; (use Pythagorean Theorem: leg^2 + leg^2 = hyp^2 )

3 0

2 years ago

A car insurance company has determined that 9% of all drivers were involved in a car accident last year. Among the 10 drivers li

____ [38]

Answer:

The correct answer: 0.05404

Step-by-step explanation:

Given:

Binomial distribution = x

n = 10

and p = 0.09

solution:

P(X=x) =1 0Cx*(0.09^x)*((1-0.09)^(10-x)) for x=0,1,2,...,10

So the probability is calculated by the Formula:

P(X>=3) = 1-P(X=0)-P(X=1)-P(X=2)

putting the given values in the formula

= 1-10C0*(0.09^0)*((1-0.09)^(10-0))-...-10C2*(0.09^2)*((1-0.09)^(10-2))

= 0.0540400

Thus, the correct answer: 0.05404

5 0

3 years ago

2) Line segment MK has endpoints at (2, 3) and (5, ?4). Segment M'K' is the reflection of MK over the y-axis. Which statement de

Marianna [84]

Answer:

Option C -M'K' is the same length as MK

Step-by-step explanation:

Given : Line segment MK has endpoints at (2, 3) and (5,4)

M'K' is the reflection of MK over the y-axis

By definition of reflection: reflection of point (x,y) across the the y-axis is the point (-x,y)

which implies M'K' has end points (-2,3) and (-5,4)

Now, we find the length of MK

let $(x_1,y_1)=(2,3)\\\\(x_2,y_2)=(5,4)$

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

⇒ $d=\sqrt{(2-5)^2+(4-3)^2}$

⇒ $d=\sqrt{9+1}$

⇒ $d=\sqrt{10}$ ....(1)

Now, we find the length of M'K'

let $(x_1^{'},y_1^{'})=(-2,3)\\\\(x_2^{'},y_2^{'})=(-5,4)$

$d^{'}=\sqrt{(x_2^{'}-x_1^{'})^2+(y_2^{'}-y_1^{'})^2}$

⇒ $d^{'}=\sqrt{(-2+5)^2+(3-4)^2}$

⇒ $d^{'}=\sqrt{9+1}$

⇒ $d^{'}=\sqrt{10}$ .....(2)

from (1) and (2) we simply show that the length of MK and M'K' is equal

we can also refer the figure attached for reflection of MK and M'K'

therefore, Option C is correct

3 0

4 years ago

Read 2 more answers

The function j(x)=39x represents the number of jumping jacks j(x) you can do in x minutes. How many jumping jacks can you do in

statuscvo [17]

The function j(x)=39x is the number of jumping jacks you can do in x minutes how many can you do in 5 minutes'?

This is asking you to evaluate the function when x is 5...so all you do here is substitute 5 for x and simplify.. since the question defines "x" as the number of minutes...and j(x) represents the number of jumping jacks.

j(x) = 39x

j(5) = 39(5)

j(5) = 195 <--- answer...

So you do 195 jumping jacks in 5 minutes

I really hope this helps :)

5 0

3 years ago

Plzzz I need help with this question I tried to solve it many times but I can't

blsea [12.9K]

Answer and Step-by-step explanation: Area of a right triangle, (as any other triangle), is calculated as: $A1=\frac{(base)(height)}{2}$

Area of a rectangle is calculated as: $A2=(side)(side)$

Area of a right trapezoid is: $A3=\frac{(a+b)h}{2}$ , where:

a is short base

b is long base

h is height

1) Expressing areas in terms of x:

Area of triangle S1:

$S1=\frac{(2x-3)(4x-6)}{2}$

$S1=4x^{2}-12x+9$

Area of rectangle S2:

$S2 = (4x-6)(3x-2)$

$S2=12x^{2}-26x+12$

Area of trapezoid S3:

$S3=\frac{(2x+3+4x+1)(2x-3)}{2}$

$S3=\frac{(6x+4)(2x-3)}{2}$

$S3=6x^{2}-5x-6$

2) a) $S=4x^{2}-12x+9+12x^{2}-26x+12-(6x^{2}-5x-6)$

$S=4x^{2}-12x+9+12x^{2}-26x+12-6x^{2}+5x+6$

$S=10x^{2}-33x+37$

Which is the same as S = (2x-3)(5x-9)

b) For the areas to be the same:

$\frac{(3x-2+3x-2+2x-3)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$\frac{(8x-7)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$32x^{2}-48x-28x+42=12x^{2}+8x-18x-12$

$20x^{2}-66x+54=0$

Using Bhaskara to solve the second degree equation:

$\frac{66+\sqrt{(-66)^{2}-(4.20.54)} }{2(20)}$

$x_{1}=\frac{66+6}{40}$ = 1.8

$x_{2}=\frac{66-6}{40}$ = 1.5

For the areas of AFGC and ADEB to be equal, x has to be 1.5 or 1.8.

c) Expand a polynomial (or equation) is to multiply all the terms, remiving the parenthesis. Reduce a polynomial (or equation) is to combine terms alike,e.g.:

$S=(2x-3)(5x-9)$