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3 years ago

6 divided by 2 (1+2)

Mathematics

2 answers:

user100 [1]3 years ago

4 0

Observe.

6/2(1+2)
6/2(3)
6/6
1

Thus, the answer is: 1

Law Incorporation [45]3 years ago

3 0

The answer is 1 if you know how to solve it

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Factorize completely the expression(h+2k)²+4k²-h²

julsineya [31]

The expression factorized completely is (h +2k)[(h+2k) + (2k-h)]

From the question,

We are to factorize the expression (h+2k)²+4k²-h² completely

The expression can be factorized as shown below

(h+2k)²+4k²-h² becomes

(h+2k)² + 2²k²-h²

(h+2k)² + (2k)²-h²

Using difference of two squares

The expression (2k)²-h² = (2k+h)(2k-h)

Then,

(h+2k)² + (2k)²-h² becomes

(h+2k)² + (2k +h)(2k-h)

This can be written as

(h+2k)² + (h +2k)(2k-h)

Now,

Factorizing, we get

(h +2k)[(h+2k) + (2k-h)]

Hence, the expression factorized completely is (h +2k)[(h+2k) + (2k-h)]

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5 0

3 years ago

Read 2 more answers

Helppp please it may be simple for u!

nikitadnepr [17]

3,1,2 replay if i got it wong

5 0

3 years ago

Write the prime factorization of each number then find the greatest common factor of the numbers 40, 48

DedPeter [7]

40= 1,2,4,5,8,10,20,40
48= 1,2,3,4,6,8,12,16,24,48
the gcf is 8.

3 0

3 years ago

Franco has $265.87 in his checking account. Then he writes the deposit slip shown. How should Franco enter the deposit in his ch

artcher [175]

Answer:

The answer is the top right one since the total $96.35 is on the deposit side.

( i did this before and got it right btw)

8 0

2 years ago

HELP PLZZ will give brainliest <3

melisa1 [442]

Answer:

Step-by-step explanation:

First question:

You are given a side, a, and its opposite angle, A. You are also given side b. Use that in the law of sines and solve for the other angle, B.

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$

$\dfrac{10}{\sin 30^\circ} = \dfrac{40}{\sin B}$

$\dfrac{1}{0.5} = \dfrac{4}{\sin B}$

$\sin B = 2$

The sine function can never equal 2, so there is no triangle in this case.

Answer: no triangle

Second question:

You are given a side, b, and its opposite angle, B. You are also given side c. Use that in the law of sines and solve for the other angle, C.

$\dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

$\dfrac{10}{\sin 63^\circ} = \dfrac{}{\sin C}$

$\sin C = \dfrac{8.9\sin 63^\circ}{10}$

$C = \sin^{-1} \dfrac{8.9\sin 63^\circ}{10}$

$C \approx 52.5^\circ$

One triangle exists for sure. Now we see if there is a second one.

Now we look at the supplement of angle C.

m<C = 52.5°

supplement of angle C: m<C' = 180° - 52.5° = 127.5°

We add the measures of angles B and the supplement of angle C:

m<B + m<C' = 63° + 127.5° = 190.5°

Since the sum of the measures of these two angles is already more than 180°, the supplement of angle C cannot be an angle of the triangle.

Answer: one triangle

3 0

3 years ago