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Nikolay [14]
1 year ago
15

Please help me I will most all info needed ! It’s due tomorrow and I’m lost and want to die.

Mathematics
1 answer:
RUDIKE [14]1 year ago
8 0

Answer:

Step-by-step explanation:

145.90

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When Eric multiplied two binomials together, his result was a trinomial. An example is
leva [86]

Answer:

(x + 4)(x - 4)

Step-by-step explanation:

There are actually quite a lot of pairs of binomials the disproves Eric's conclusion, but they all model after the same special product: a^2 - b^2.

The special product a^2 - b^2 can be factored into (a + b)(a - b) and for all real a and b, it will come out as a binomial.

Here is an example:

(x + 4)(x - 4)

We can use the distributive property to get:

x^2 - 4x + 4x - 16

which is the same as

x^2 - 16

This would disprove Eric's conclusion.

6 0
3 years ago
Bobby can run a mile in 7 minutes. Use the ratio table to help determine how
meriva

Answer:

4 miles

Step-by-step explanation:

In 7 minutes he runs 1 mile

So in 28 minutes he runs 4 miles

<u>How did I get this answer?</u>

7×4=28

So 1×4=4

In ratios, when you times anything by something, you must do the same to the other part of the ratio.

7 0
2 years ago
Read 2 more answers
(3Q) Evaluate the logarithm.
Aleonysh [2.5K]

Answer:

a. 5/3

that's your answer

3 0
3 years ago
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Need help please assist me find the area of a regular hexagon
garri49 [273]

Answer:

The area of the regular hexagon is 166.3\ units^{2}

Step-by-step explanation:

we know that

The area of a regular hexagon can be divided into 6 equilateral triangles

so

step 1

Find the area of one equilateral triangle

A=\frac{1}{2}(b)(h)

we have

b=r=8\ units

h=4\sqrt{3}\ units ----> is the apothem

substitute

A=\frac{1}{2}(8)(4\sqrt{3})

A=16\sqrt{3}\ units^{2}

step 2

Find the area of 6 equilateral triangles

A=(6)16\sqrt{3}=96\sqrt{3}=166.3\ units^{2}

7 0
3 years ago
Writeanequationforthelinethatpassesthrough(4,-3)and has a slope of 3.
djyliett [7]

Answer:

x+y=1 (if x=4 )

4+y=1

y=1-4

y=-3

7 0
2 years ago
Read 2 more answers
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