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Olenka [21]
3 years ago
5

HELP ME SOLVE THE AREA OF THIS TRAINGLE

Mathematics
1 answer:
alexdok [17]3 years ago
7 0
A triangle is 1/2 bh so .5*8*6 which equals 25 cm square. hope I helped. brainliest would be much appreciated because I am trying to rank up!!!
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Plzzzzz help will give brain and points!!! Explain why a quadratic equation with a positive discriminant has two real solutions,
NISA [10]

Answer:

The given equation is -2x^2 = -8x + 8

This can be rewritten as 2x^2 - 8x + 8 = 0

Here the value of a =2, b = -8 and c = 8

Discriminant = b^2 - 4ac

Now plug in the above values in the discriminant, we get

= (-8)^2 - 4*2*8

= 64 - 64

= 0

Here the discriminant is 0, so we will get one real root.

The discriminant is equal to 0, which means the equation has one real number solution.

Hope this will helpful.

Thank you.

5 0
4 years ago
Penny's garden has alength of 16ft. and the width is 5 ft. with a rectangular fountain in the middle that measures 5 ft by 9 ft.
lilavasa [31]

First we need the whole garden:

16 x 5 = 80

Now we need to fountain:

5 x 9 = 45

Now to find our answer:

80 - 45 = 35

So our answer is B. 35ft²

4 0
3 years ago
Sofia is learning to knit. After her first three stitches, she had to undo the last two. Then she made three more stitches and a
Marizza181 [45]
She’ll have to make 4 more stitches in order to get 6 good ones. Each time she creates 3 she undoes 2 which after 4 more times would leave her with 6 good stitches.
4 0
3 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
2 years ago
Please help! What is 1 1/2+1/6?<br><br> -giving out 20 points
Mice21 [21]

1  \frac{2}{3}

Hope it helps

7 0
3 years ago
Read 2 more answers
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