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3 years ago

RATIOS MY WORST ENEMY HELP!!! IM SO LOST

Mathematics

1 answer:

Pavel [41]3 years ago

4 0

(4,6) (6,9) (8,12)
Now for rick it is (6,8) (9,12) (12,16)

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You are competing in a race. The table shows the times from last year's race you want your time to be last year's median time wi

aleksandr82 [10.1K]

Answer:

$27\leq x \leq37$

Step-by-step explanation:

Let's arrange the numbers from least to greatest:

20, 25, 27, 28, 31, 32, 32, 38, 40, 40, 42, 48

The median is the "middle" number when arranged from least to greatest.

There are 12 numbers, so the middle number would be "average" of 6th and 7th terms.

6th term = 32

7th term = 32

Average = (32+32)/2 = 32

Hence, median = 32

Absolute deviation of 5 minutes means 5 minutes less or greater than the median, 32.

So, 32 - 5 = 27 & 32 + 5 = 37

Hence $27\leq x \leq37$

4 0

2 years ago

The side of an Equileteral triangle is 12cm. What is its Area?

elena55 [62]

Answer:

A = 62.35 cm²

Step-by-step explanation:

Use the area formula A = $\frac{\sqrt{3}a^2}{4}$ , where a is the side length.

Plug in the values:

A = $\frac{\sqrt{3}(12^2)}{4}$

A = $\frac{\sqrt{3}(144)}{4}$

A = 62.35 cm²

3 0

3 years ago

I got 40% off when I purchased a set of golf balls regularly priced at $20. How<br> much did I save?

lianna [129]

Answer:

Step-by-step explanation:

40% times $20 = 8

BRAINLIEST PLEASE

6 0

3 years ago

Can you make more arrays with 8 tiles then 9

Charra [1.4K]

No because u have to have 9 or 9 or 8 and 8

5 0

3 years ago

Read 2 more answers

Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge

Ede4ka [16]

Answer:

Part a) The quadratic function is $4x^{2} +20x-39=0$

Part b) The value of x is $1.5\ in$

Part c) The photo and frame together are $7\ in$ wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

$(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0$

Part b) What is the value of x?

Solve the quadratic equation $4x^{2} +20x-39=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem

we have

$4x^{2} +20x-39=0$

$a=4\\b=20\\c=-39$

substitute in the formula

$x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}$

$x=\frac{-20(+/-)\sqrt{1,024}} {8}$

$x=\frac{-20(+/-)32} {8}$

$x=\frac{-20(+)32} {8}=1.5\ in$ -----> the solution

$x=\frac{-20(-)32} {8}=-6.5\ in$

Part c) How wide are the photo and frame together?

$(4+2x)=4+2(1.5)=7\ in$

5 0

3 years ago