How many different 4-question geometry quizzes can a teacher make if there are 7 different problems to test on?.

Mathematics

1 answer:

stich3 [128]3 years ago

3 0

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.

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Which value of b would make x²+bx-30 factorable?<br> A. -31<br> B. -17<br> C. 11<br> D. 13

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Answer:

D. 13

Step-by-step explanation:

well since it's a quadratic equation, it can be written in the form of ax²+bx+c=0. So we have a=1, b=b and c= -30. Also the roots are given by the equation-

x= -b± $\sqrt{b^{2}-4ac}$ /2a. So all I did was put the number from the choices which would satisfy the sqrt part and that turned out to be 13 since $\sqrt{13^{2}-4*1*(-30)$ = $\sqrt{289}$ = 17. So b=13.