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kolezko [41]
3 years ago
11

Evaluate the expression 9 - z9−z9, minus, z when z = 4z=4z, equals, 4

Mathematics
1 answer:
Sophie [7]3 years ago
7 0

Answer:

Step-by-step explanation:

Answer:

<h2>5</h2>

Step-by-step explanation:

Given the expression f(z) = 9-z

To get the value of the resulting value of the function when z = 4, all we need to do is substitute the value of z given into the function to have;

f(4) = 9-4

f(4) = 5

The right answer is 5

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The safety instructions for a 20 foot ladder say the ladder should not be inclined more than 70 degrees with the ground. suppose
Romashka [77]

Answer: the distance of the base of the house to the foot of the ladder is 6.84 feet

Step-by-step explanation:

The scenario is shown in the attached photo.

Right angle triangle ABC is formed when the ladder leans against the wall of the house.

AC = the height of the ladder

AB = x feet = distance of the base of the house to the foot of the ladder

BC is the wall of the building.

To determine x, we will apply trigonometric ratio

Cos # = adjacent/hypotenuse

Where

# = 70 degrees

Hypotenuse = 20

Adjacent = x

Cos 70 = x/20

x = 20cos70

x = 20 × 0.3420

x = 6.84 feets

4 0
3 years ago
What makes the statement true? 7^-2=
Damm [24]
  <span>What makes the statement true?</span>7^-2= 0.0204081633
8 0
3 years ago
Read 2 more answers
Find the slope of a line from these two coordinates.
tankabanditka [31]

This is wrong.

It's supposed to be -4/15.

But try -4/5.

7 0
3 years ago
Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
I don't understand any of these questions
Xelga [282]
20 because there are many more
6 0
3 years ago
Read 2 more answers
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