X
у
Which equation could be used to create the data shown in the
table?
1
5
O y = 5x
4
17
O y = 5x + 1
5
21
O y = 4x + 1
7
29
O y = x + 4
9
37

Answer:

1/4

Step-by-step explanation:

White = w

Black = B

Blue = b1

Tan = t

Wb1

Wt

Bbi

Bt

The answer will be 1/4, because there are 4 ways it can work and only 1 way it can be white shirt and tan pants.

Answer:

1/4

Step-by-step explanation:

it would be 1/4 because there are 4 different clothing pieces in total and there is only one way it would work the way the problem says.

Step-by-step explanation:

[tex]\tan \alpha + \cot\alpha = \dfrac{\sin \alpha}{\cos \alpha} +\dfrac{\cos \alpha}{\sin \alpha}[/tex]

[tex]=\dfrac{\sin^2\alpha + \cos^2\alpha}{\sin\alpha\cos\alpha}=\dfrac{1}{\sin\alpha\cos\alpha}[/tex]

[tex]=\left(\dfrac{1}{\sin\alpha}\right)\!\left(\dfrac{1}{\cos\alpha}\right)=\csc \alpha \sec\alpha[/tex]

Question :

Required solution :

Here we would be considering L.H.S. and solving.

Identities as we know that,

By using the identities we gets,

[tex] : \: \implies \: \sf{ \dfrac{sin \: \alpha }{cos \: \alpha} \: + \: \dfrac{cos \: \alpha }{sin \: \alpha} }[/tex]

[tex]: \: \implies \: \sf{ \dfrac{sin \: \alpha \times sin \: \alpha }{cos \: \alpha \times sin \: \alpha} \: + \: \dfrac{cos \: \alpha \times cos \: \alpha }{sin \: \alpha \times \: cos \: \alpha } } [/tex]

[tex] : \: \implies \: \sf{ \dfrac{sin {}^{2} \: \alpha }{cos \: \alpha \times sin \alpha} \: + \: \dfrac{cos {}^{2} \: \alpha }{sin \: \alpha \times \: cos \: \alpha } } [/tex]

[tex]: \: \implies \: \sf{ \dfrac{sin {}^{2} \: \alpha }{cos \: \alpha \: sin \alpha} \: + \: \dfrac{cos {}^{2} \: \alpha }{sin \: \alpha \: cos \: \alpha } } [/tex]

[tex]: \: \implies \: \sf{ \dfrac{sin {}^{2} \: \alpha \: + \: cos {}^{2} \alpha}{cos \: \alpha \: sin \alpha} } [/tex]

Now, here we would be using the identity of square relations.

By using the identity we gets,

[tex] : \: \implies \: \sf{ \dfrac{1}{cos \: \alpha \: sin \alpha} }[/tex]

[tex]: \: \implies \: \sf{ \dfrac{1}{cos \: \alpha } \: + \: \dfrac{1}{sin\: \alpha} }[/tex]

[tex]: \: \implies \: \bf{sec \alpha \: cosec \: \alpha}[/tex]

Answer:

48, 6

Step-by-step explanation:

The ratio of the pieces is 6 to 1

Add them together to get the total

6+1 = 7

Divide the total length by 7

56/7 = 8

Multiply the ratios by 8

6*8 = 48

 1*8 = 6

The peices are 48 and 6

Answer:

b. The actual count of bike riders is too small.

d. n*p is not greater than 10.

Step-by-step explanation:

Confidence interval for a proportion:

To be possible to build a confidence interval for a proportion, the sample needs to have at least 10 successes, that is, [tex]np \geq 10[/tex] and at least 10 failures, that is, [tex]n(1-p) \geq 10[/tex]

Of the 123 students surveyed 5 ride a bike to campus.

Less than 10 successes, that is:

The actual count of bike riders is too small, or [tex]np < 10[/tex], and thus, options b and d are correct.

(4) y = (3 - 2x)³ + 24x

Use the power and chain rules:

dy/dx = 3 (3 - 2x)² d/dx [3 - 2x] + 24

dy/dx = 3 (3 - 2x)² (-2) + 24

dy/dx = -24x ² + 72x - 30

(5) y = 54x - (2x - 7)³

Same basic procedure:

dy/dx = 54 - 3 (2x - 7)² d/dx [2x - 7]

dy/dx = 54 - 3 (2x - 7)² (2)

dy/dx = -24x ² + 168x - 240

Answer:

30 clocks

Step-by-step explanation:

Set up an equation:

Variable x = number of clocks

1200 + 20x = 60x

Isolate variable x:

1200 = 60x - 20x

1200 = 40x

Divide both sides by 40:

30 = x

Check your work:

1200 + 20(30) = 60(30)

1200 + 600 = 1800

1800 = 1800

Correct!

Answer:

4 e

Step-by-step explanation:

dz6dxrx xrrx6 xz33x4xr4x xrx

Answer:

4x+ 4

Step-by-step explanation:

Answer:

3:8

Step-by-step explanation:

i will gadit

that only

Answer:

1/425

Step-by-step explanation:

The first card can be any card, so we don’t have to evaluate the probability.

Now we can suppose that the exit card is a two

- For the second card we have 3/51 of possibilities that is a 2 = 1/17

- For the third card we have 2/50 of possibilities that is a 2 = 1/25

1/17 * 1/25 = 1/425

Answer:

Hello

Step-by-step explanation:

The domain is limited with 2 lines parallel: -1 ≤ x ≤ 1

and the disk ? (inside of a circle) of center (0,0) and radius 2

[tex]dom\ f(x,y)=\{(x,y) \in \mathbb{R} ^2 | \ -1\leq x \leq -1\ and \ ( -\sqrt{4-x^2} \leq \ y \leq \sqrt{4-x^2}\ ) \ \}\\[/tex]

Answer:

55000×100/90

61,111.111

Answer:

No

Step-by-step explanation:

If we use the Pythagorean theorem, we can find if it is a right triangle. To do that, set up an equation.

[tex]6^{2}+8^{2}=c^2[/tex]

If the triangle is a right triangle, c would equal 11

Solve.

[tex]36+64=100[/tex]

Then find the square root of 100.

The square root of 100 is 10, not 11.

So this is not a right triangle.

I hope this helps!

Answer:

5

Step-by-step explanation:

(625 ^2)^(1/8)

Rewriting 625 as 5^4

(5^4 ^2)^(1/8)

We know that a^b^c = a^(b*c)

5^(4*2)^1/8

5^8 ^1/8

5^(8*1/8)

5^1

5

Answer:

[tex]5[/tex]

Step-by-step explanation:

[tex] { {(625}^{2} )}^{ \frac{1}{8} } \\ { ({25}^{2 \times 2} )}^{ \frac{1}{8} } \\ {25}^{4 \times \frac{1}{8} } \\ {5}^{2 \times 4 \times \frac{1}{8} } \\ {5}^{ \frac{8}{8} } \\ {5}^{1} \\ = 5[/tex]

Answer:

khai niem hinh cat don gian?

Answer:

0.5679

Step-by-step explanation:

From. The table Given above :

The probability of female Given an advanced degree ;

P(F|A) = p(FnA) / p(A)

From the table, p(FnA) = 322

P(Advanced degree), P(A) = (245 + 322) = 567

Hence,

P(F|A) = p(FnA) / p(A) = 322 / 567 = 0.5679

Answer:

3.1 gallons

Step-by-step explanation:

To solve this, we need to figure out how many gallons of gas go into 72 miles. We know 23 miles is equal to one gallon of gas, and given that the ratio of miles to gas stays the same, we can say that

miles of gas / gallons = miles of gas / gallons

23 miles / 1 gallon = 72 miles / gallons needed to go to Bob's mother's house

If we write the gallons needed to go to Bob's mother's house as g, we can say

23 miles / 1 gallon = 72 miles/g

multiply both sides by 1 gallon to remove a denominator

23 miles = 72 miles * 1 gallon /g

multiply both sides by g to remove the other denominator

23 miles * g = 72 miles * 1 gallon

divide both sides by 23 miles to isolate the g

g = 72 miles * 1 gallon/23 miles

= 72 / 23 gallons

≈ 3.1 gallons

Answer:

10:3

Step-by-step explanation:

Make 2 1/2 an improper fraction, you will get 5/2. You dont have to do anything to the 3/4.

For you to find the ratio of an fraction, you have to take the numerator but the denominator has to be the same.

So make 5/2 to a 10/4.

Take the numerator 10 & 3.

Your answer will be 10:3

No problem.

9514 1404 393

Answer:

  x = 7

  y = 5

Step-by-step explanation:

The applicable rule of exponents is ...

  a^-b = 1/a^b

__

For a=-j and b=7,

  (-j)^-7 = 1/(-j)^7   ⇒   x = 7

For a=k and b=-5,

  k^-5 = 1/k^5   ⇒   y = 5

Answer:

5 ≤ x ≤ 10             5 ≤ y ≥ -1

Step-by-step explanation:

Answer:

Scatter plot charts are good for relationships and distributions, but pie charts should be used only for simple compositions — never for comparisons or distributions.

(1) Both equations in (a) and (b) are separable.

(a)

[tex]\dfrac xy y' = \dfrac{2y^2+1}{x+1} \implies \dfrac{\mathrm dy}{y(2y^2+1)} = \dfrac{\mathrm dx}{x(x+1)}[/tex]

Expand both sides into partial fractions.

[tex]\left(\dfrac1y - \dfrac{2y}{2y^2+1}\right)\,\mathrm dy = \left(\dfrac1x - \dfrac1{x+1}\right)\,\mathrm dx[/tex]

Integrate both sides:

[tex]\ln|y| - \dfrac12 \ln\left(2y^2+1\right) = \ln|x| - \ln|x+1| + C[/tex]

[tex]\ln\left|\dfrac y{\sqrt{2y^2+1}}\right| = \ln\left|\dfrac x{x+1}\right| + C[/tex]

[tex]\dfrac y{\sqrt{2y^2+1}} = \dfrac{Cx}{x+1}[/tex]

[tex]\boxed{\dfrac{y^2}{2y^2+1} = \dfrac{Cx^2}{(x+1)^2}}[/tex]

(You could solve for y explicitly, but that's just more work.)

(b)

[tex]e^{x+y}y' = 3x \implies e^y\,\mathrm dy = 3xe^{-x}\,\mathrm dx[/tex]

Integrate both sides:

[tex]e^y = -3e^{-x}(x+1) + C[/tex]

[tex]\ln(e^y) = \ln\left(C - 3e^{-x}(x+1)\right)[/tex]

[tex]\boxed{y = \ln\left(C - 3e^{-x}(x+1)\right)}[/tex]

(2)

(a)

[tex]y' + \sec(x)y = \cos(x)[/tex]

Multiply both sides by an integrating factor, sec(x) + tan(x) :

[tex](\sec(x)+\tan(x))y' + \sec(x) (\sec(x) + \tan(x)) y = \cos(x) (\sec(x) + \tan(x))[/tex]

[tex](\sec(x)+\tan(x))y' + (\sec^2(x) + \sec(x)\tan(x)) y = 1 + \sin(x)[/tex]

[tex]\bigg((\sec(x)+\tan(x))y\bigg)' = 1 + \sin(x)[/tex]

Integrate both sides and solve for y :

[tex](\sec(x)+\tan(x))y = x - \cos(x) + C[/tex]

[tex]y=\dfrac{x-\cos(x) + C}{\sec(x) + \tan(x)}[/tex]

[tex]\boxed{y=\dfrac{(x+C)\cos(x) - \cos^2(x)}{1+\sin(x)}}[/tex]

(b)

[tex]y' + y = \dfrac{e^x-e^{-x}}2[/tex]

(Note that the right side is also written as sinh(x).)

Multiply both sides by e ˣ :

[tex]e^x y' + e^x y = \dfrac{e^{2x}-1}2[/tex]

[tex]\left(e^xy\right)' = \dfrac{e^{2x}-1}2[/tex]

Integrate both sides and solve for y :

[tex]e^xy = \dfrac{e^{2x}-2x}4 + C[/tex]

[tex]\boxed{y=\dfrac{e^x-2xe^{-x}}4 + Ce^{-x}}[/tex]

(c) I've covered this in an earlier question of yours.

(d)

[tex]y'=\dfrac y{x+y}[/tex]

Multiply through the right side by x/x :

[tex]y' = \dfrac{\dfrac yx}{1+\dfrac yx}[/tex]

Substitute y(x) = x v(x), so that y' = xv' + v, and the DE becomes separable:

[tex]xv' + v = \dfrac{v}{1+v}[/tex]

[tex]xv' = -\dfrac{v^2}{1+v}[/tex]

[tex]\dfrac{1+v}{v^2}\,\mathrm dv = -\dfrac{\mathrm dx}x[/tex]

[tex]-\dfrac1v + \ln|v| = -\ln|x| + C[/tex]

[tex]\ln\left|\dfrac yx\right| -\dfrac xy = C - \ln|x|[/tex]

[tex]\ln|y| - \ln|x|  -\dfrac xy = C - \ln|x|[/tex]

[tex]\boxed{\ln|y| -\dfrac xy = C}[/tex]

Answer:

270 degrees

Step-by-step explanation:

If you plug in 270 in place of the x's, the function is true!

This is correct for Plate/Edmentum users!! Hope I could help :)

Answer:

B

Step-by-step explanation:

substitute the values in the eq. Ot is also arithmetic progression.

If 6ˣ = 5, then

(6ˣ)³ = 6³ˣ = 5³ = 125,

and

6³ˣ⁺² = 6³ˣ × 6² = 125 × 6² = 125 × 36 = 4500

Given:

The logarithmic equation is:

[tex]\log 970=x[/tex]

To find:

The exponential equations that is equivalent to the given logarithmic equation.

Solution:

Property of logarithm:

If [tex]\log_b a=x[/tex], then [tex]a=b^x[/tex]

We know that the base log is always 10 if it is not mentioned.

If [tex]\log a=x[/tex], then [tex]a=10^x[/tex]

We have,

[tex]\log 970=x[/tex]

Here, base is 10 and the value of a is 970. By using the properties of exponents, we get

[tex]970=10^x[/tex]

Interchange the sides, we get

[tex]10^x=970[/tex]

Therefore, the correct option is B, i.e., [tex]10^x=970[/tex].

Note: It should be "=" instead of "-" in option B.

Answer:

8

Step-by-step explanation:

As it's an isosceles right triangle, it's sides are equal, say x. x^2+x^2=(4*sqrt(2))^2. x=4, Area is (4*4)/2=8

Step-by-step explanation:

AB = square root of [(xA-xB)^2+(yA-yB)^2]

AB=Squarerootof(-1-11)^2 +(-3-(-8))^2=Squarerootof(-12)^2+(5)^2)

AB=Squarerootof((144)+25)= Squarerootof(169)=13 the answer is 13 units

The choice D is the right one