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3 years ago

Beginning at sunrise the temperature rose 2.8° per hour how long did it take for the temperature to rise a total of 19.18°?

Mathematics

2 answers:

Viktor [21]3 years ago

7 0

What Temperature Was It When It Began?

nadezda [96]3 years ago

3 0

19.18/2.8= 6.85 hours or rounded it would be 7 hours

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Hi I need help with this packet because my teacher explains it very hardly for me

algol13

Answer:

(-1,4) (-8,2) (-6,10)

Step-by-step explanation:

All you had to do is reflect it and figure out the new coordinates. There is you answer.... you welcome

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3 years ago

Jenny thinks that the expression below is equal to x²-4. If you agree, show that she is correct. If you disagree, show that she

sergiy2304 [10]

Answer:

Jenny is wrong the correct answer is x²+ 4 - 4 x.

Step-by-step explanation:

Given that

(x-2)³∕x-2

(x-2)³∕(x-2) =(x-2) (x-2)²∕(x-2)

(x-2)³∕(x-2) = (x-2)²

As we know that

(a-b)² = a²+b² - 2 ab

(x-2)² = x²+ 2² - 2 ˣ 2 ˣ x = x²+ 4 - 4 x

(x-2)² = x²+ 4 - 4 x

It means that

(x-2)³∕(x-2) = (x-2)² = x²+ 4 - 4 x

So Jenny is wrong the correct answer is x²+ 4 - 4 x.

8 0

3 years ago

What is the interquartile range of the following data set? 3,8, 14, 19, 22, 29, 33, 37, 43, 49. A. 26 B. 23 C. 19 D. 18

Fynjy0 [20]

Answer:

I think the answer is B.23

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3 years ago

In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is th

olya-2409 [2.1K]

Answer:

$\frac{BE}{EC} =\frac{1}{3}$

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

$\frac{BK}{CK} =\frac{1}{4}$

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK and Δ EBK are similar

so we have

$\frac{AD}{BE} =\frac{DK}{BK}$

$\frac{AD}{BE} =\frac{4}{1}$

$\frac{BC}{BE}=\frac{4}{1}$ ( ABCD is parallelogram so AD=BC)

$\frac{BE+EC}{BE}=\frac{4}{1}$ ( BC= BE+EC)

$\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}$

$1+\frac{EC}{BE}=4$

$\frac{EC}{BE}=3$ ( subtracting 1 from both side )

$\frac{EC}{BE}=\frac{3}{1}$

taking reciprocal both side

$\frac{BE}{EC} =\frac{1}{3}$

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3 years ago

When you graph a system of equations what are the three possible outcomes?

siniylev [52]

Answer:

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