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anzhelika [568]
3 years ago
13

What is the measure of angle A?

Mathematics
2 answers:
maksim [4K]3 years ago
7 0

Answer:

I remember this when I was in grade 5 the answer is 250 now I am in college

Oxana [17]3 years ago
3 0

Answer: 110

Step-by-step explanation:

I guessed

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Using the discriminant, how many real number solutions does this equation have? 3x^2 – 2 = 5x
Anvisha [2.4K]
The discriminant of this equation is 49. The solutions for this equation are x=2 and x=-\frac{1}{3}
I hope that helps.
7 0
3 years ago
For thanksgiving, you made 32 cupcakes. How many cupcakes were leftover if your family ate 3/4 of them?
Anvisha [2.4K]
All you need to do is divide 32 by 4.

32/4= 8

So, 3*8= 24 (that is 3/4 of the cupcakes)

32-24= 8

8 is the answer.

I hope this helps!

Brainliest answer is always appreciated! 
4 0
3 years ago
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What is the answer for x+11+8x=29 has to show work
m_a_m_a [10]

Answer:

x=2

Step-by-step explanation:

9x+11=29

9x=29-11

9x=18

x=2

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3 years ago
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PLEASE HELP IN IMAGE nonsense will be reported
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The answer is -2. The line is going from left to right which makes it negative. Then if you look at the line and do change in y(rise) over change in x(run).
8 0
4 years ago
How many committees of 4 boys and 3 girls<br> can be formed from a class of 6 boys and 7<br> girls?
VLD [36.1K]

Answer:

525

Step-by-step explanation:

This is a question involving combinatorics

The number of ways of choosing a subset k from a set of n elements is given by {n \choose k} which evaluates to \frac{n!}{k!(n-k)!}

n! is the product n × (n-1) × (n-2) x....x 3 x 2 x 1

For example,

4! = 4 x 3 x 2 x 1 = 24

3! = 3 x 2 x 1 = 6

Since we have to choose 4 boys from a class of 6 boys, the total number of ways this can be done is

{6 \choose 4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!}

Note that 6! = 6 x 5 x 4 x 3 x 2 x 1 and 4 x 3 x 2 x 1  is nothing but 4!

So the numerator can be re-written as 6 x 5 x (4!)

We can rewrite the expression \frac{6!}{4!2!} \text{ as } \frac{6.5.4!}{4!2!}

Cancelling 4! from both numerator and denominator gives us the result

as  (6 × 5)/2! = 20/2 = 15 different ways of choosing 4 boys from a class of 6 boys

For the girls, the number of ways of choosing 3 girls from a class of 7 girls is given by

{7 \choose 3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!}

This works out to (7 x 6 x 5 )/(3 x 2 x 1)  (using the same logic as for the boys computation)

= 210/6 = 35

So total number of committees of 4 boys and 3 girls that can be formed from a class of 6 boys and 7 girls = 15 x 35 = 525

8 0
2 years ago
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