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rusak2 [61]
3 years ago
13

A one-celled organism measures 32 millimeters in length in a photograph. If the photo has been enlarged by a factor of 100, what

is the actual length of the organism?
Mathematics
2 answers:
tatyana61 [14]3 years ago
7 0
The answer is 0.32 millimeters - move the decimal point over the amount of zeroes.
Hope this helped! Rate brainliest if you want. :)
Have a nice day!

maw [93]3 years ago
4 0
Let x mm be the original size of the cell.

After enlarging by 100 times, the cell length becomes 100.x

Or 100x = 32 mm, then x= 32/100 →and x = 0.32 mm which the original size

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What is the image point of (-1,4) after a translation left 5 units and up 2 units
shepuryov [24]

Answer:

(-6,6)

Step-by-step explanation:

8 0
3 years ago
Classify the following triangle
larisa86 [58]

Answer:

A. Right Triangle

D. Isosceles

Step-by-step explanation:

Well, since it has a square at the bottom of the triangle, it means it's a right triangle, because it has a 90 degree angle.

It is also an isosceles triangle, because two of the sides are of equal length.

Hope this helped!

:)

4 0
3 years ago
About 245 out of 500 students prefer walking instead of riding the bus to school. About what percent of the students prefer walk
nasty-shy [4]

Answer:

49 percent

Step-by-step explanation:

So convert the fraction into a percentage

245/500= 49%

7 0
3 years ago
What is 10a - 5b if a = 25 and b = 3​
AURORKA [14]

Answer:

\boxed{235}

Step-by-step explanation:

<em>Hey there!</em>

Well given that,

a = 25

b = 3

10(25) - 5(3)

250 - 15

= 235

<em>Hope this helps :)</em>

6 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

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
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