Expand the function.
f(x)
=
(2x + 3)4
16x4 +
[? ]x3
+
]x2
+
x +
IX +
х

Answer:

I understand how that feels

Step-by-step explanation:

9514 1404 393

Answer:

  A. {2x, x<1; (x+5)/2, x≥1}

Step-by-step explanation:

The left side of the graph (x<1) has positive slope and a y-intercept of 0. Only one answer choice matches: choice A.

Answer:

False

Step-by-step explanation:

The sides are not equal.

Answer:

chu papi

Step-by-step explanation:

Using the probability concept, it is found that there is a [tex]\mathbf{\frac{1}{k}}[/tex] probability that “Relaxation Through Mathematics” is first to review.

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem:

The probability is:

[tex]p = \frac{D}{T} = \frac{1}{k}[/tex]

A similar problem is given at https://brainly.com/question/24483829

Answer:

29.5

Step-by-step explanation:

Multiply 606 and 20.5 and get 29.5.

Answer:

You told me to resolve this question

Answer:

LJ< JK<LK

Step-by-step explanation:

for a theorem, the order of the side of the triangle is the same as the order of the opposite angles since for the angles it's K<L<J, you replace each with the opposite side (quick trick, it's the one not containing the letter!)

Answer:

A child psychologist is interested in the number of times a newborn baby’s crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let X = the number of times per week a newborn baby’s crying wakes its mother after midnight. For this example, x = 0, 1, 2, 3, 4, 5.

P(x) = probability that X takes on a value x.

x P(x)0 P(x = 0) = 250

1 P(x = 1) = 1150

2 P(x = 2) = 2350

3 P(x = 3) = 950

4 P(x = 4) = 450

5 P(x = 5) = 150

Step-by-step explanation:

pa follow po pls.

Answer:

0.00706666666

Step-by-step explanation:

Answer:

No

No

Yes

Yes

No

Step-by-step explanation:

In this case we are looking at substitution, so any where you find x put three(3) in replace of it.

And remember the answer should be 36,

I find it difficult to type the answer so I will take a picture instead.

Answer:

36

Step-by-step explanation:

The LCM of 4, 9, and 12 is the product of all prime numbers on the left, i.e. LCM(4, 9, 12) by division method = 2 × 2 × 3 × 3 = 36.

[tex]\frac{4}{9} -\frac{1}{12} = \frac{4.4}{9.4} - \frac{1.3}{12.3} = \frac{16}{36}- \frac{3}{36} = \frac{13}{36}[/tex]

ok done. Thank to me :>

Answer:

four times a number less 10 is equall to 16 ia not

If the given equation is

[tex]27^{\frac13} = 9^{14} + 3^{x + 1}[/tex]

first simplify the left side using the fact that 27 = 3³:

[tex]27^{\frac13} = (3^3)^{\frac13} = 3^{\frac33} = 3^1 = 3[/tex]

and on the right side, 9 = 3²:

[tex]9^{14} = (3^2)^{14} = 3^{2\times14} = 3^{28}[/tex]

So we have

[tex]3 = 3^{28} + 3^{x + 1}[/tex]

Next,

[tex]3^{x+1} = 3^x \times 3^1 = 3 \times 3^x[/tex]

so that

[tex]3 = 3^{28} + 3 \cdot 3^x[/tex]

and dividing both side by 3 gives

[tex]1 = 3^{27} + 3^x[/tex]

Isolate 3ˣ :

[tex]3^x = 1 - 3^{27}[/tex]

Solve for x by taking the base-3 logarithm of both sides:

[tex]\log_3(3^x) = \log_3(1-3^{27})[/tex]

[tex]x \log_3(3) = \log_3(1-3^{27})[/tex]

[tex]\boxed{x = \log_3(1-3^{27})}[/tex]

Answer:

0.09

Step-by-step explanation:

Answer:

W=20 V=42 X=87

Step-by-step explanation:

all sides are equal meaning that two corners from across each other are equal so v is 42 and w 20 this also means x is 87 im not too sure about y

Answer:

No

Step-by-step explanation:

While these values are close, they are not equal, so I know the original fractions cannot be proportional to each other. So my answer is: The fractions are not

Answer:

-3

Step-by-step explanation:

a. Simply put (x+h) wherever you have x.

[tex]f(x+h) = -3(x+h)+7 = -3x +7 -3h[/tex]

b. We have the first part, just subtract

[tex]f(x+h) = -3(x+h)+7 = -3x +7 -3h - (-3x+7) = -3x+7-3h +3x -7 = -3h[/tex]

c. Finally, we divide by h[tex]\frac{-3h}h = -3[/tex]

Answer:

66

forty-four times 1.5 equals 66

Answer:

No, sorry about that, buddy, I cannot.

Answer:

-0.366666666667

Step-by-step explanation:

that's what you get when you divide -22/5 (-4 2/5) by 4/3 (1 1/3)

Answer:

-24/5 is the correct answer.

Steps:

1) -4 x 2/5 = -8/5

2) 1 x 1/3 = 1/3

3) -8/5 : 1/3 = -24/5

Step-by-step explanation:

each child gets a certain amount of items costing the specified price by item.

so, we need to multiply the number of children with the number of items with the price per item to get the total cost.

500×3×3.5 for pencils = 5,250

500×5×7.5 for notebooks = 18,750

500×1×21.75 for crayons = 10,875

500×2×9.26 for pad paper = 9,260

total 44,135

since we have only 1,200 available in the existing fund, we need a new one to cover the remaining

44,135 - 1,200 = 42,935

Answer:

6 four-packs.

Step-by-step explanation:

My reasoning for 6 four-packs is that, the shopper needs 24, right? And if that shopper buys 6 four-packs it'll be cheaper than the larger package. The 6 four-packs come to a cost of $20.04. But the 4 six-packs will come out as $21.12. The shopper will be able to save $1.08 :)

The measures of the angles and the lengths of the sides of the triangle can

be found using sine rule.

The correct responses are;

Reasons:

The given parameter are;

∠B = 65°20', b = 5.38 inches, c = 4.88 inches

∠C ≥ 90°

By sine rule, we have;

[tex]\dfrac{5.38}{sin(65^{\circ} 20')} = \mathbf{ \dfrac{4.88}{sin(\angle C)}}[/tex]

[tex]sin(\angle C) = \mathbf{ \dfrac{sin(65^{\circ} 20') \times 4.88}{5.38}}[/tex]

[tex]\angle C = arcsin \left(\dfrac{sin(65^{\circ} 20') \times 4.88}{5.38} \right) \approx \mathbf{ 55.517 ^{\circ}}[/tex]

∠C = 55.517° < 90°, therefore, the correct response is; None

[tex]\dfrac{sin(65^{\circ} 20')}{5.38} = \dfrac{4.88}{sin(\angle C)}[/tex]

∠A = 180° - 65°20' - 55.517° = 59.15° = 59°9'

∠A = 59°9'

Therefore;

[tex]\dfrac{5.38}{sin(65^{\circ} 20')} = \mathbf{ \dfrac{a}{sin(59.15^{})}}[/tex]

[tex]a = \dfrac{5.38 \times sin(59.15^{\circ}) }{sin(65^{\circ} 20')} \approx \mathbf{ 5.083}[/tex]

a ≈ 5.083 inches

Learn more here:

https://brainly.com/question/15018190

Let

[tex]\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots[/tex]

Differentiating twice gives

[tex]\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots[/tex]

[tex]\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n[/tex]

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

[tex]\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0[/tex]

[tex]\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0[/tex]

Then the coefficients in the power series solution are governed by the recurrence relation,

[tex]\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}[/tex]

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

[tex]k=0 \implies n=0 \implies a_0 = a_0[/tex]

[tex]k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}[/tex]

[tex]k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}[/tex]

[tex]k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}[/tex]

It should be easy enough to see that

[tex]a_{n=2k} = \dfrac{a_0}{(2k)!}[/tex]

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

[tex]k = 0 \implies n=1 \implies a_1 = a_1[/tex]

[tex]k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}[/tex]

[tex]k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}[/tex]

[tex]k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}[/tex]

so that

[tex]a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}[/tex]

So, the overall series solution is

[tex]\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)[/tex]

[tex]\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}[/tex]