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3 years ago

Based on the graph, what is one solution of the equation f(x) = g(x)?

Mathematics

1 answer:

OLEGan [10]3 years ago

6 0

Answer:

x=1

x=-5

Step-by-step explanation:

we know that

The solution of the system of equations is the intersection point both graphs

In this problem there are two intersection points

The system has two solutions

we have

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

$g(x)=x^{2} +3x-4$

one solution is the point (-4.9, 5.4) and the other (0.92, -0.42)

see the attached figure

therefore

x=1

x=-5

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Write the number five hundred two thousand, six hundred seventy-eight in standard form.

Arada [10]

The answer is 502678

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2 years ago

What is the arc length of the arc sub tended in a circle with radius 6 and an angle of 7 pi/ 8

KengaRu [80]

Answer:

Formula for the Arc length is given by:

$\text{Arc length} = 2 \pi r \cdot \frac{\theta}{360^{\circ}}$

As per the statement:

radius of circle(r) = 6 units

Angle ( $\theta$ ) = $\frac{7 \pi}{8}$ radian

Use conversion:

$1 rad = \frac{180}{\pi}$

$\frac{7 \pi}{8}$ = $\frac{180}{\pi} \cdot \frac{7 \pi}{8} = \frac{1260}{8} = 157.5^{\circ}$

then;

substitute these given values we have;

Use value of $\pi = 3.14$

$\text{Arc length} = 2\cdot 3.14 \cdot (6) \cdot \frac{157.5^{\circ}}{360^{\circ}}$

$\text{Arc length} = 2\cdot 3.14 \cdot (6) \cdot 0.4375$

Simplify:

$\text{Arc length}=16.485$

Therefore, the arc length of the arc substended in a circle with radius 6 units an angle of 7 pi/8 is 16.485 units

7 0

3 years ago

How does the volume of an oblique cylinder change if the radius is reduced to 2/9 of its original size and the height is quadrup

vesna_86 [32]

Volume of a cylinder = π r² h

Let us assume the following values:
radius = 9
height = 10

Volume = 3.14 * 9² * 10
= 3.14 * 81 *10
Volume = 2,543.40

Changes:
radius is reduced to 2/9 of its original size = 9 x 2/9 = 2
height is quadrupled = 10 x 4 = 40

Volume = π r² h
= 3.14 * 2² * 40
= 3.14 * 4 * 40
Volume = 502.40

Original volume = 2543.40 V.S. Volume after change = 502.40

The volume of an oblique cylinder decreased when its radius was decreased to 2/9 of its original size and its height is increased 4 times.

4 0

2 years ago

Let f(x)=g(x)h(x), g(10)=-4, h(10)=560, g'(10)=0, and h'(10)=35. find f'(10)

Ulleksa [173]

This question has extraneous info to trick you.
f(x) = g(x)h(x) ⇒ f'(x) = g'(x)h'(x) . Letting x = 10, we get f'(10) = g'(10)h'(10). then just plug in the values provided. g(10) and h(10) are there to throw you off, just use g'(10) and h'(10).
So f'(10), pronounced "eff prime of ten", = 0 * 35 = 0.

If the question were asking for f(10) instead of f'(10) then you would use g(10) and h(10), ⇒-4*560=90.

7 0

3 years ago

HELPPPP FIND THE AREA

GuDViN [60]

you can just split the figure in 2 separate figures and them add the areas:

area rec. 1:

3m×3m= 9m²

area rec. 2:

3m× (2m+3m)

= 3m×5m

=15m²

total area:

9m²+15m²= 24m²

5 0

1 year ago

Read 2 more answers