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Rudiy27
3 years ago
15

Does 369 + 306 equal to 675 or 678

Mathematics
2 answers:
Elden [556K]3 years ago
4 0

Answer:675

Step-by-step explanation:

mr Goodwill [35]3 years ago
3 0

375 is the correct answer homito

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Possible values of x in 6x^2 + 432 = 0
Wewaii [24]
6x^2 + 432 = 0 \\
x^2+72=0\\
x^2=-72\\
x=-\sqrt{-72} \vee x=\sqrt{-72}\\
x=-6\sqrt2 i \vee x=6\sqrt 2 i

3 0
4 years ago
Given the m<8=45, find the other angle measures. Be able to say how you found each angle measure. PLEASE help!
iris [78.8K]

Answer:

m<1 = 135°, m<2 = 45°, m<3 = 135°, m<4 = 45°, m<5 = 135°, m<6 = 45°, m<7 = 135°

Step-by-step explanation:

We know that a straight line always gives us a measure of 180° total. This would mean 180° = m<8 + m<7. So, if we plug in the real value of m<8, we get 180 = 45 + m<7. From there, we can subtract 180 by 45, and we get 135° = m<7.

We know that vertical angles are congruent - so m<5 is the same as m<7 - making m<5 = 135° as well. This could also apply to m<8 and m<6, so m<6= 45°.

From there, we can also say that alternate interior angles are congruent to each other - meaning m<5 = m<3, and m<6 = m<4. So, m<3 = 135° and m<4 = 45°.

Alternate exterior angles are congruent too, which means m<8 = m<2, and m<7 = m<1. So, m<2 = 45° and m<1 = 135°.

In summary,

m<7 = 135° because angle subtraction.

m<5 = 135° and m<6= 45° because vertical angles are congruent.

m<3 = 135° and m<4 = 45° because alternate interior angles are congruent.

m<1 = 135° and m<2 = 45° because alternate exterior angles are congruent!

Hope this makes sense! Not sure if I explained it well.

3 0
4 years ago
At noon a tank contains 6 cm of water after several hours it contains 3 cm of water what is the percent inecrease of water in th
Hitman42 [59]

Question:

At noon a tank contains 6 cm of water after several hours it contains 3 cm of water what is the percent decrease of water in the tank

Answer:

The percent Decrease of water in the tank  is 50%

Step-by-step explanation:

Given:

Initial amount of water in the tank at the noon = 6 cm

amount of water remained in water after several hours = 3 cm

To Find:

The percent decrease =?

Solution:

The percent decrease in the tank

=>\frac{\text {water present after some hours }}{\text {initial water contained in the tank }} \times 100

Substituting the values

=>\frac{3}{6} \times 100

=>\frac{1}{2} \times 100

=>\frac{100}{2}

=> 50

5 0
3 years ago
You are 5 feet tall and cast an 8-foot shadow. A land post nearby casts a shadow that is 20 feet. Which equation can you use to
Anuta_ua [19.1K]
5ft. x
—- = —-
8ft. 20ft
6 0
4 years ago
Read 2 more answers
PLZ HELP ASAP IT IS TIMED
Oduvanchick [21]

Answer:

The first one

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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