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Harlamova29_29 [7]

3 years ago

15

What is an equation of the line, in point-slope form, that passes through the given point and has the given slope? point:(11, 3)

; slope:4/11

1 answer:

Galina-37 [17]3 years ago

8 0

Y=4/11x-1

:c) :c) :c)
-Oliver

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What are the zeros of the quadratic function f(x) = 2x2 + 16x – 9?

Aleksandr-060686 [28]

Answer:

The zeros of the function are -0.488 and -7.39

Step-by-step explanation:

Given the quadratic function

f(x) = 2x²+16x-9

The zeros of the function is gotten at when f(x) = 0

2x²+16x-9 = 0

Using the general equation

x = -b±√b²-4ac/2a

From the equation, a = 2, b = 16 and c= -9

x = -16±√16²-4(2)(-9)/2(2)

x = -16±√256-72/4

x = -16±√184/4

x = -16+13.56/4

x = -16+13.56/4 and -16.13.56/4

x = --2.44/5 and -29.56/4

x = -0.488 and -7.39

8 0

2 years ago

Ms Quon discovered that 40 students like soccer, 20 liked baseball, 10 liked football, and 5 liked swimming.

LiRa [457]

$1.\ 20:10=2\\2\ many\ times\ of\ students\ liked\ baseball\ as\ football\\\\2.\ 40:10=4\\4\ many\ times\ of\ students\ liked\ soccer\ as\ football\\\\3.\ There\ is\ no\ information\ about\ basketball\ students$

7 0

3 years ago

Trig please help

enot [183]

Answer:

One

Step-by-step explanation:

9/sin(34) = 6/sinC

sinC = 0.372795269

C = 21.9, 158.1

Since 158.1 + 34 = 192.1 > 180

It will not form a triangle

So,

Only one triangle with angle 21.9° at C

5 0

3 years ago

Which pair of triangles can be proven by the HL theorem

natulia [17]

Also known as the hypotenuse leg theorem. It states that if the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent.' This is kind of like the SAS, or side-angle-side postulate. Congruent triangles.

4 0

3 years ago

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$

Hence, the correct answer is option A) 1260

5 0

3 years ago