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3 years ago

14

Find the limit of the function by using direct substitution. limit as x approaches four of quantity x squared plus three x minus

one

2 answers:

MAXImum [283]3 years ago

8 0

Answer:

The correct answer is 27, when using direct substitution

Step-by-step explanation:

(4)^2+3(4)-1 --> 16+12-1 --> 16+11= 27

Yuliya22 [10]3 years ago

5 0

$\lim_{x \to 4} f(x)=x^2+3x-1$
just subsitute
f(4)=4²+3(4)-1
f(4)=16+12-1
f(4)=28-1
f(4)=27

it approaches 27 as x approaches 4

You might be interested in

Cos(−θ)=√3/3, sinθ<0

Delicious77 [7]

Answer:

$\sin\theta=-\frac{\sqrt{6}}{3}$

Step-by-step explanation:

The given trigonometric equation is $\cos(-\theta)=\frac{\sqrt{3} }{3}$ .

We can either use the Pythagorean identity or the right angle triangle to solve for $\sin\theta$ .

According to the Pythagorean identity,

$\cos^2\theta+\sin^2\theta=1$

Recall that, the cosine function is an even function, therefore

$\cos(-\theta)=\cos(\theta)$

$\Rightarrow \cos(\theta)=\frac{\sqrt{3} }{3}$ .

We substitute this value in to the above Pythagorean identity to get;

$(\frac{\sqrt{3}}{3})^2+\sin^2\theta=1$

$\Rightarrow \frac{3}{9}+\sin^2\theta=1$

$\Rightarrow \sin^2\theta=1-\frac{3}{9}$

$\Rightarrow \sin^2\theta=\frac{6}{9}$

$\Rightarrow \sin\theta=\pm \sqrt{\frac{6}{9}}$

$\Rightarrow \sin\theta=\pm \frac{\sqrt{6}}{3}$

But we were given that,

$\sin\theta\:$ , so we choose the negative value.

$\Rightarrow \sin\theta=-\frac{\sqrt{6}}{3}$

The correct answer is B

7 0

3 years ago

Please help it’s due tonight. Your help will be greatly appreciated.There is also a graph. WILL PUT YOU AS BRAINLY!!!!!!

chubhunter [2.5K]

Answer: see below

Step-by-step explanation:

The shape is a trapezoid

Base 1 (BH) is 10

Base 2 (QA) is 14

Height is 5

Area

$A=\frac{base_{1} +base_{2} }{2} h$

$A=\frac{10+14}{2} *5\\A=12*5\\A=60$

Find distance between QB and HA

$D=\sqrt{(x_{2} -x_{1}) ^{2}+(y_{2}-y_{1}) ^{2}$

Q( -4,4) B (-2,9)

$D=\sqrt{(-2--4)^{2}+(9-4)^{2} } \\D=\sqrt{2^{2} +5^{2} } \\D=\sqrt{29}$

Perimeter

$P=10+14+2\sqrt{29} \\P=24 +2\sqrt{29} \\P= 24+10.77\\P=34.77$

5 0

3 years ago

Which function has only one x-intercept at (-6, 0)?

Lelu [443]

Answer:

<h2>The function f(x) = (x - 6)(x - 6) has only one x-intercept. But at (6, 0) not at (-6, 0).</h2>

Step-by-step explanation:

The intercept form of a quadratic equation (parabola):

$y=a(x-p)(x-q)$

p, q - x-intercepts

Therefore

The function f(x) = x(x - 6) = (x - 0)(x - 6) has two x-intercepts at (0, 0) and (6, 0)

The function f(x) = (x - 6)(x - 6) has only one x-intercept at (6, 0)

The function f(x) = (x + 6)(x - 6) = (x - (-6))(x - 6)

has two x-intercept at (-6, 0) and (6, 0)

The function f(x) = (x + 1)(x + 6) = (x - (-1))(x - (-6))

has two x-intercepts at (-1, 0) and (-6, 0).

6 0

3 years ago

Read 2 more answers

In 16years, Ben will be 3 times as old as he is right now.

dedylja [7]

Answer:

Don't quote me on this but I believe the answer is that he is 8 years old right now if that's what you're looking for

Step-by-step explanation:

8 + 16 = 24

8 × 3 = 24

That's what I thought

5 0

3 years ago

PLEASE HELP!<br>Calculate $40^{13} \pmod{85}.$

SVEN [57.7K]

In this type of calculations, we decompose 13 by checking the lowest powers of the base, that is 40. for example we check 40^2, or 40^3 and compare it to 85

Notice

40*40*40=64,000

so we check how many time does 85 fit into 64,000:

64,000/85=752.94

85*753=64,005; 64000-64,005=-5

this means that

$40^{3} =-5\pmod{85}$

thus

$40^{13} =40^{3*4+1}={(40^{3})}^{4}*40=(-5)^{4}*40 \pmod{85}=\\\\625*40\pmod{85}=(7*85+30)*40\pmod{85}=30*40\pmod{85}\\\\=1200\pmod{85}=(14*85+10)\pmod{85}=10\pmod{85}$

Answer: 10 (mod85)

Remark, the set of all solutions is:

{......-75, 10, 95, .....}, that is 85k +10

7 0

3 years ago