Ask question

Ask question

All categories

3 years ago

12

Given a right triangle, find the measures of all the angles, in degrees, if one angle is a right angle and the measurement of th

e second angle is six less than seven times the measure of the third angle.

1 answer:

vivado [14]3 years ago

3 0

Give the known angle (90) a pronumeral of r.
give the second angle (the one with all the subtraction) a pronumeral or x.
give the third angle the pronumeral of y

x = 7y - 6
90 + x + y = 180
90 + 7y - 6 + y = 180
90 + 8y - 6 = 180
90 + 8y =186
8y = 96
y = 12. the third angle is 12

so 12 + 90 = 102
the second angle must equal 88 to make it to 180

You might be interested in

7 less or greater than 5x -16 does it equal 4?

Marysya12 [62]

Can you re-phrase the question

4 0

3 years ago

5x-(x+3)=1/3 (9x+18)-5<br> Please show me step by step

zzz [600]

Step-by-step explanation:

5x-(x+3)=1/3(9x+18)-5

5x-x+3=3x+6-5

1/3 multiplied by 9/1 is three because if you take away the ones and put your problem like this: 9/3, you'll get 3.

1/3 multiplied by 18/1 is six because if you take away the ones and switch to division, it'll look like this: 18/3. 18/3 is 6.

5x-1x+3=3x+6-5

If a variable is alone, you should - in this case - put a one before it, but if the variable is alone you should know that it's automatically a one.

4x+3=3x-1

6-5=1 Simple math problem, probably self explanatory

4x+3=3x-1

Subtract 3x from 4x and you'll get 1x because 4-3 is 1.

1x+3=-1

Then you add 3 to -1 which is 2.

The answer is 1x=2

Or you could've done:

4x+3=3x-1

-4x -4x

3=-1x-1

+-1 +-1

2=1x

It's the same answer, but all I did was subtract the 4x from the 4x and 3x.

3 0

3 years ago

If the ratio of boys to girls at the school is 2:5 and there are 40 boys how many girls are there?

sukhopar [10]

Answer:

100 girls

Step-by-step explanation:

The ratio of boys : girls is 2 : 5.

In other words, boys are 2 parts of the school while girls are 5 parts of the school.

Since there are 40 boys, 2 parts of the school is 40. Thus:

$2p = 40\\p = 20$

1 part is 20.

Since there are 5 parts girls, there are

$5p = 5*20 = 100$

100 girls.

Answer: 100 girls

5 0

3 years ago

Bartholemew will draw with replacement 160 tickets at random from a box with five tickets [0, 0, 0, 1, 2]. Estimate the chance t

cupoosta [38]

Answer:

0.0786

Step-by-step explanation:

It is given that Bartholemew had drawn the replacement of 160 tickets.

There are five tickets = [0, 0, 0, 1, 2]

Now we need to find the estimate of the ticket that has 1 on it and it turns up on the 32 draws exactly.

Since the probability of the drawing 1 out of 5 tickets is given by, $\frac{1}{5} = 0.2$

So the binomial with the parameter of n = 160 and p = 0.2, we get

P (it turns up on exactly 32 draws) = P(X = 32)

Therefore,

$C(160, 32)\times (0.2)^{32}\times (0.8)^{160-32}$

= 0.0786

4 0

3 years ago

Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se

defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

$p_{1}(t) = 1$ , $p_{2}(t)= t^{2}$ and $p_{3}(t) = 3 + 3\cdot t$ :

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1} + 3\cdot \alpha_{3} = 0$

$\alpha_{2} = 0$

$\alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

$p_{1}(t) = t$ , $p_{2}(t) = t^{2}$ and $p_{3}(t) = 2\cdot t + 3\cdot t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0$

$(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0$

The following system of linear equations is obtained:

$\alpha_{1}+2\cdot \alpha_{3} = 0$

$\alpha_{2}+3\cdot \alpha_{3} = 0$

Since the number of variables is greater than the number of equations, let suppose that $\alpha_{3} = k$ , where $k\in\mathbb{R}$ . Then, the following relationships are consequently found:

$\alpha_{1} = -2\cdot \alpha_{3}$

$\alpha_{1} = -2\cdot k$

$\alpha_{2}= -2\cdot \alpha_{3}$

$\alpha_{2} = -3\cdot k$

It is evident that $\alpha_{1}$ and $\alpha_{2}$ are multiples of $\alpha_{3}$ , which means that the set of vector are linearly dependent.

$p_{1}(t) = 1$ , $p_{2}(t)=t^{2}$ and $p_{3}(t) = 3+3\cdot t +t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2}+ \alpha_{3}\cdot (3+3\cdot t+t^{2}) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1+(\alpha_{2}+\alpha_{3})\cdot t^{2}+3\cdot \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1}+3\cdot \alpha_{3} = 0$

$\alpha_{2} + \alpha_{3} = 0$

$3\cdot \alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

The set of vectors A and C are linearly independent.

4 0

3 years ago