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4 years ago

If f and g are continuous functions with f(3)=5 and lim x-->3 [2f(x)-g(x)]= 4, find g(3)

1 answer:

7 0

Since $f$ is continuous, you know that

$\displaystyle\lim_{x\to3}f(x)=f(3)=5$

So,

$\displaystyle\lim_{x\to3}(2f(x)-g(x))=2f(3)-\lim_{x\to3}g(x)=10-g(3)=4$

where the limit for $g$ also is equal to the value of $g(3)$ because $g$ is also continuous. This means $g(3)=6$ .

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Complete the statement using always, sometimes, or never. A rectangle is __________________ a rhombus.

Lesechka [4]

Hi!

A rectangle can sometimes be a rhombus if it has sides of all equal lengths. If a rectangle was a rhombus, it would also be a square. Furthermore, a square is always a rhombus.

Hope this helps! :)

7 0

3 years ago

State the order and type of each transformation of the graph of the function

algol [13]

Using translation concepts, compared to the graph of the base function g(x) = x², the function f(x) was:

Vertically stretched by a factor of 4.

Shifted left by 9 units.

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

In this problem, the base function g(x) = x² was:

Multiplied by 4, that is, vertically stretched by a factor of 4.

Shifted left 9 units, as x -> x + 9.

More can be learned about translation concepts at brainly.com/question/4521517

#SPJ1

6 0

2 years ago

Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$

3 0

3 years ago

Read 2 more answers

16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.

Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$

3 0

3 years ago

Round 58.6790691589 to the nearest hundred-thousandth.

suter [353]

Answer:

58.678

Step-by-step explanation:

Just make the answer reach 3 numbers after the decimal point.

6 0

3 years ago

Read 2 more answers