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2 years ago

5

Megan has been planting young trees in her garden. One maple tree is 66 centimeters tall and is growing 8 centimeters per month.

One oak tree is 70 centimeters tall and is growing 7 centimeters per month.

1 answer:

EastWind [94]2 years ago

7 0

Once per month there is food that is 33 centimeters long

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I need to find the median ,mean,mode,range,and

Makovka662 [10]

Median: 23.5
12 and 35 are in the middle so you add them and divide by 2, leaving you
23.5 as your median.
Mean: 41.3
Add all numbers together and divide the sum by the number of numbers added. So all numbers added together equals 413. Divide by ten the nuber of numbers added. 413/10 equals 41.3.
Mode: 99
The number that appears most often. In this case, 99 would be the mode.
Range: 98
The lowest number subtracted from the highest number. 99-1 equals 98.
IQR: 84.75

Hope this helped! :)

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Algebraic factorisation simplification long question

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Step-by-step explanation:

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3 years ago

The bacteria took 1 hour and 40 minutes to fill the initial bottle. Explain how it only took 10 minutes to fill 3 additional emp

geniusboy [140]

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3 years ago

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Solve the equation for all values of x by completing the square.<br> x2 + 8.0 + 15 = 0

Aleonysh [2.5K]

Answer:

x= -3

x= -5

Step-by-step explanation:

x^2 + 8x + 15 = 0

find which 2 numbers when multiplied equal 15 and when added equal 8

they are 3 and 5

(x+3)(x+5)

x+3 =0

x+5=0

x= -3

x= -5

6 0

3 years ago

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

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

so

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

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