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Helga [31]
3 years ago
14

PLEASE HELP ME OUT PLEASE

Mathematics
1 answer:
Yuliya22 [10]3 years ago
5 0

Answer:

Can't help if I cant see the entire question, sry :(

Step-by-step explanation:

You might be interested in
find the two angles measures from the diagram 7x+9 5x+3 there's 3 answers 1. 73 and 73 2. 107 ,107 3.107 and 73 this is so confu
viva [34]

Answer:

<h2>107 and 73</h2><h2 />

Step-by-step explanation:

supplementary angle

7x + 9 + 5x + 3 = 180

12x = 180 - 3 - 9

x = 168 / 12

x = 14

plugin x=14 into the equation

= 7x + 9

= 7(14) + 9

= 107°

plugin x=14 into the equation

= 5x + 3

= 5(14) + 3

= 73°

4 0
3 years ago
How many different 4-question geometry quizzes can a teacher make if there are 7 different problems to test on?.
stich3 [128]

Answer:

35 quizzes

Step-by-step explanation:

We need to determine how many ways we can choose 4 questions out of 7 to make the quizzes.

We do this by using the formula \displaystyle nCr=\frac{n!}{r!(n-r)!} which describes the number of ways we can choose r objects given n possible choices. So, if n=7 and r=4, then:

_4C_7=\frac{7!}{4!(7-4)!}\\\\_4C_7=\frac{7*6*5*4!}{4!*3!}\\\\_4C_7=\frac{210}{6}\\\\_4C_7=35

Hence, the teacher can make 35 different geometry quizzes.

3 0
3 years ago
Can you please solve this!!
lyudmila [28]

Answer: A

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Use the quadratic formula to solve the following equation -3x^2-x-3=0
tigry1 [53]

<u>Answer:</u>

x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i are two roots of equation -3 x^{2}-x-3=0

<u>Solution:</u>

Need to solve given equation using quadratic formula.

-3 x^{2}-x-3=0

General form of quadratic equation is a x^{2}+b x+c=0

And quadratic formula for getting roots of quadratic equation is

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}

Since b^{2}-4 a c is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}

8 0
4 years ago
What is the verbal expression 21-18
valina [46]

Answer: twenty one decreased by eighteen or the difference of twenty one and eighteen

5 0
3 years ago
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