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iren [92.7K]
3 years ago
5

Kanye wants to make a triangular flower bed using logs with the lengths 3 ft, 4 ft, and 8ft. Can Kanye form a triangle with the

logs without cutting any of them? Explain.
Mathematics
1 answer:
valina [46]3 years ago
5 0
Way 1:
a triangle with sides 3,4,5 will make a triangle. because this isn't 3,4,5 triangle it wont work.



way 2:
By using Heron's formula you can find the area of the flower bed.
<span>s=(a+b+c) ÷2     
</span><span>Area =√s(s-a)(s-b)(s-c)
</span>so, 3+4+8= 15
15÷2 =7.5
√7.5(7.5-3)(7.5-4)(<span>7.5-8)
</span>Now because 7.5-8 gives you a negative number and you can't have a negative number under a square root unless you want to use an imaginary number, it won't form a triangle.
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aivan3 [116]

Answer:

B

Step-by-step explanation:

If Sean wants to but both the muffin and cookie then this addition. 3 +m shows the correct expression. the answer is b

4 0
3 years ago
Can anyone help me with this question? Best answer will get brainliest!
Svet_ta [14]
Ion know. What’s the question
8 0
2 years ago
What does it mean in Math when we say that an equation is balanced?
tekilochka [14]
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7 0
3 years ago
Looking to solve (Ac+b-d)(a^2-c)=
Sunny_sXe [5.5K]

It should be noted that the expansion of the equation (ac+b-d)(a² - c) will be a³c + ac² + a²b - bc - a²d + cd.

<h3>How to illustrate the information?</h3>

It should be noted that an equation is used to show the relationship that occur between the variables that are illustrated.

In this case, it should be noted that the equation

(ac+b-d)(a² - c) will be solved accordingly. This will be:

(ac+b-d)(a² - c)

Open the parentheses

a³c + ac² + a²b - bc - a²d + cd

Therefore, it should be noted that the expansion of the equation (ac+b-d)(a² - c) will be a³c + ac² + a²b - bc - a²d + cd.

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6 0
1 year ago
Write in standard form 2+4i over 1+i
Levart [38]
So hmm a conjugate, is pretty much just, the same binomial, but with a different sign in the middle, so, a + b, has a conjugate of a - b
or -a + b, has a conjugate of - a  - b, or c - d, has a conjugate of c +d, and so on

anyway, the idea being, to "rationalize" the expression, namely, getting rid of the pesky radical in the denominator

so, we'll multiply the expression by 1, since anything times 1 is just itself

however, bear in mind, that 1, can be a/a, or b/b, or cheese/cheese, or anything/anything

so, we'll multiply the top and bottom of the fraction, by the conjugate of the denominator

anyhow, that said   \bf \cfrac{2+4i}{1+i}\cdot \cfrac{1-i}{1-i}\implies \cfrac{(2+4i)(1-i)}{(1+i)(1-i)}\implies \cfrac{2-2i+4i-4i^2}{1-i^2}\\\\&#10;-----------------------------\\\\&#10;\textit{recall that }i^2=-1\\\\&#10;-----------------------------\\\\&#10; \cfrac{2-2i+4i-4(-1)}{1-(-1)}\implies \cfrac{2+2i+4}{2}\implies \cfrac{6+2i}{2}\implies \cfrac{6}{2}+\cfrac{2i}{2}&#10;\\\\\\&#10;\boxed{3+i}
7 0
3 years ago
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