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Karolina [17]
3 years ago
13

Help Please !!!! Very Confused, do not understand how to complete these 2 questions properly.

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
8 0

Answer:

11) tan(7x+3) = OE/OJ

7x+3=arctan(OE/OJ)=45

x=6


12) a) x = 40

<ajq = <acq (base angles of isosceles triangle)

50+x+90=180

x=40


b) they are congruent triangle, reason HL   (hypotenuse, leg)

aj=ac

aq=qa

right angle equal

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Graph the following lines and write the equation in slope-intercept form.
Taya2010 [7]

Answer:

y=mx+b

3=slope

1=y coordinate

-8=x coordinate

b=y intercept

Substitute m for 3

y=3x+b

Next, substitute the coordinate values

1=3(-8)+b

Solve

1=-24+b

+24 +24

25=b

Now substitute b and m for their respective values

y=3x+25

Step-by-step explanation:

3 0
2 years ago
Find the product of -2x a(-4x b - 2x 3 + 5x)
Anastasy [175]

Answer:

        2                   2

8abx  +12ax-10ax

Step-by-step explanation:

(the twos are exponents)

5 0
2 years ago
The function g(x) is a transformation of the quadratic parent function, fx):
77julia77 [94]

Answer: D

Step-by-step explanation:

The graph is narrower, so the coefficient of x^2 has an absolute value more than 1.

  • Eliminate B and C.

Also, because g(x) opens down, the coefficient of x^2 is negative.

  • Eliminate A.

This leaves D as the answer.

6 0
1 year ago
Factor completely.
Serga [27]
Factor the polynomial:

4u² – 20u + 25

Rewrite – 20u as – 10u – 10u, and then factor it by grouping:

= 4u² – 10u – 10u + 25

= 2u * (2u – 5) – 5 * (2u – 5)

= (2u – 5) * (2u – 5)

= (2u – 5)² <––– this is the answer.

I hope this helps. =)
5 0
3 years ago
You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.
postnew [5]

<em><u>This problem seems to be wrong because no minimum point was found and no point of landing exists</u></em>

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

<u>Derivatives</u>

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

1)

The function provided in the question is

h(x)=2x^2-12x+20

Let's find the first derivative

h'(x)=4x-12

solving h'=0:

4x-12=0

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

2)

To find when will the ball land, we set h=0

2x^2-12x+20=0

Simplifying by 2

x^2-6x+10=0

Completing squares

x^2-6x+9+10-9=0

Factoring and rearranging

(x-3)^2=-1

There is no real value of x to solve the above equation, so the ball will never land.

This problem seems to be wrong because no minimum point was found and no point of landing exists

3 0
3 years ago
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