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sasho [114]
3 years ago
9

Given g(x)= -x-2, find g(0)

Mathematics
1 answer:
german3 years ago
8 0

Answer:

-2

Step-by-step explanation:

You just have to plug in 0 for x

-(0)-2

there's no such thing as a negative 0 so it would just be negative 2

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What is the quotient of 2+3x2-47-12) - (x2 + 5x+8)
kap26 [50]
The quotient to this answer is -x^2-5x-59
4 0
3 years ago
What is the solution of the equation 1.6m - 4.8 = -1.6m
Flauer [41]
Solve for m:
1.6 m - 4.8 = -1.6 m

Add 1.6 m to both sides:
1.6 m + 1.6 m - 4.8 = 1.6 m - 1.6 m

1.6 m - 1.6 m = 0:
1.6 m + 1.6 m - 4.8 = 0

1.6 m + 1.6 m = 3.2 m:
3.2 m - 4.8 = 0

Add 4.8 to both sides:
3.2 m + (-4.8 + 4.8) = 4.8

4.8 - 4.8 = 0:
3.2 m = 4.8

Divide both sides of 3.2 m = 4.8 by 3.2:
(3.2 m)/3.2 = 4.8/3.2

3.2/3.2 = 1:
m = 4.8/3.2

4.8/3.2 ≈ 1.5:
Answer:  m ≈ 1.5
6 0
4 years ago
Help asap-3(1+6j)=14-j
Elina [12.6K]

Answer:

J= -1

Step-by-step explanation:

5 0
3 years ago
Given f(x), find g(x) and h(x) such that f(x)=g(h(x)) and neither g(x) nor h(x) is solely x. f(x)=−3x+4⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3+5
iVinArrow [24]

Answer:

h(x)=-2x+4

g(x)=\sqrt[3]{x}-5

Step-by-step explanation:

Consider the given function is

f(x)=\sqrt[3]{-2x+4}-5

It is given that f(x)=g(h(x)) and neither g(x) nor h(x) is solely x.

f(x)=\sqrt[3]{(-2x+4)}-5

Let h(x)=-2x+4, then we get

f(x)=g(h(x))=\sqrt[3]{h(x)}-5

Substitute h(x)=x in the above function.

g(x)=\sqrt[3]{x}-5

Therefore, the required functions are h(x)=-2x+4 and g(x)=\sqrt[3]{x}-5.

Check the solutions.

g(h(x))=g(-2x+4)             [\because h(x)=-2x+4]

g(h(x))=\sqrt[3]{-2x+4}-5            [\because g(x)=\sqrt[3]{x}-5]

g(h(x))=f(x)

Therefore, our solution is correct.

7 0
4 years ago
In trapezoid $PQRS,$ $\overline{PQ} \parallel \overline{RS}$. Let $X$ be the intersection of diagonals $\overline{PR}$ and $\ove
jeyben [28]

The area of the trapezoidal comes out to be 125 square units.

<h3>What is trapezoidal?</h3>

A four-sided shape called a trapezoid has one pair of parallel sides. It fundamentally resembles a square, rectangle, or parallelogram in two dimensions.

Now, according to the question;

If a trapezoid PQRS of parallel sides PQ & RS has always been divided into 4 triangles by it's own diagonals PR and QS, which intersect at X,

Then, the area for triangle PSX equals which of triangle QRX, as well as the product of triangle PSX but also triangle QRX equals the same of triangle PQX & triangle RSX.

Let area of the triangle PSX is A₁

Let area of the triangle QRX is A₂

Let area of the triangle PQX is A₃

Let area of the triangle RSX is A₄

Then,

A₁ × A₂ = A₃ × A₄

As A₁ = A₂

The,

A₁² = A₃ × A₄

The areas for the triangles;

A₃  = 20 square units

A₄ = 45 square units.

Substitute the values;

A₁² = 20 × 45

      = 900

A₁ = 30

Thus, A₁ = A₂ = 30 square units.

The entire area of the trapezium is;

= A₁ + A₂ + A₃ + A₄

= 30 + 30 + 20 + 45

= 125

Therefore, the complete area of the trapezium PQRS is 125 square units.

To know more about the trapezoidal, here

brainly.com/question/1410008

#SPJ4

The correct question is-

In trapezoid PQRS, PQ is parallel to RS. Let X be the intersection of diagonals PR and QS. The area of triangle PQX is 20 and the area of triangle RSX is 45. Find the area of trapezoid PQRS.

4 0
1 year ago
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