what is the least common multiple between 25 and 8

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

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

Answer:

False

Step-by-step explanation:

well i think that the answer from my calculations

Answer: A (x+1)(x+2)(x+5)

Step-by-step explanation:

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

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:

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:

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.

If 6ˣ = 5, then

(6ˣ)³ = 6³ˣ = 5³ = 125,

and

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

Answer:

B

Step-by-step explanation:

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

"RS tangent to circle a..." is first statement          Reason: Given

Second Reason: "Radius perpendicular to tangent"

Second Statement: "AR is parrallel to BS"      Reason: "2 lines perpendicular..."

Answer:

khai niem hinh cat don gian?

Answer:

Angle A = Angle D = 69° 30'

Angle B = Angle C = 110° 30'

Step-by-step explanation:

     B ___ C

    /              \

  /                  \

A ________ D

AB and CD are 10

BC is 6

AD is 14

If we divide the trapezoid, we can imagine a line.

     B_ F_C

    /      |      \

  /        |         \

A ___E____ D

AE = ED = 7 (14/2)

BF = FC = 3

So now, we draw another line from B or C to AE or ED

     B_ F_ C

    /      |     | \

  /        |     |    \

A ___E_ G_ D

EG = GD = 3.5 (7/2)

There is a right triangle now, GCD

GD is 3.5 and CD is 10. To determine angle D, we can apply trigonometric function:

CD is H, and GD is A

cos D = A/H

cos D = 3.5/10 → 0.35

angle D = 69° 30'

By theory, we know that angle D and angle A, are the same so:

Angle D = Angle A = 69° 30'

Angle B = Angle C

We also make a cuadrilateral, which is EFCD.

Angle D is 69° 30', Angle E is 90°, Angle F is also 90°

Sum of angles in cuadrilateral is 360°

360° - 69° 30' - 90° - 90° = Angle C = Angle B

Angle C = Angle B = 110° 30'

Let's confirm the angles in the trapezoid:

69° 30' + 110° 30' + 69° 30' + 110° 30' = 360°

A + B + C + D

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.

Answer:

5 ≤ x ≤ 10             5 ≤ y ≥ -1

Step-by-step explanation:

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:

106 people.

Step-by-step explanation:

Logistic equation:

The logistic equation is given by:

[tex]P(t) = \frac{K}{1+Ae^{-kt}}[/tex]

In which

[tex]A = \frac{K - P_0}{P_0}[/tex]

K is the carrying capacity, k is the growth/decay rate, t is the time and P_0 is the initial value.

Suppose a rumor is going around a group of 191 people. Initially, only 38 members of the group have heard the rumor.

This means that [tex]K = 191, P_0 = 38[/tex], so:

[tex]A = \frac{191 - 38}{38} = 4.03[/tex]

Then

[tex]P(t) = \frac{191}{1+4.03e^{-kt}}[/tex]

3 days later 68 people have heard it.

This means that [tex]P(3) = 68[/tex]. We use this to find k.

[tex]P(t) = \frac{191}{1+4.03e^{-kt}}[/tex]

[tex]68 = \frac{191}{1+4.03e^{-3k}}[/tex]

[tex]68 + 274.04e^{-3k} = 191[/tex]

[tex]e^{-3k} = \frac{191-68}{274.04}[/tex]

[tex]e^{-3k} = 0.4484[/tex]

[tex]\ln{e^{-3k}} = \ln{0.4484}[/tex]

[tex]-3k = \ln{0.4484}[/tex]

[tex]k = -\frac{\ln{0.4484}}{3}[/tex]

[tex]k = 0.2674[/tex]

Then

[tex]P(t) = \frac{191}{1+4.03e^{-0.2674t}}[/tex]

How many people are expected to have heard the rumor after 6 days total have passed since it was initially spread?

This is P(6). So

[tex]P(6) = \frac{191}{1+4.03e^{-0.2674*6}} = 105.52[/tex]

Rounding to the nearest whole number, 106 people.

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:

Z =  -1.60

it is low ... it appears that for this problem 2 standard deviations below must be reached to be considered "unusual"

Step-by-step explanation:

Answer:

4x+ 4

Step-by-step explanation:

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

Answer:

answer D

Step-by-step explanation:

V=L*W*H=1 ==> L=1,W=1,H=1

A:

L-> L+2=1+2=3

W -> W+2 = 1+2=3

H -> H+2=1+2=3

V=3*3*3=27 not the doubled of the volume's cube

A is false

B:

H -> H+2=1+2=3

V=1*1*3=3 not the doubled of the volume's cube

B is false

C:

H -> 2*H=2*1=2

L -> 2*L=2*1=2

W -> 2*W = 2*1=2

V=2*2*2=8 not the doubled of the volume's cube

C is false

D:

H-> H*2=1*2=2

L=1

W=1

V=1*1*2=2 is the doubled of the volume's cube

D is true

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.

Answer:

it's. is now the MA plz I miss you

si pudiera escoger entre vivir eternamente y vivir dos veces

yo escogeria vivir dos veces porque vivir una vida eterna sin ti a mi lado seria el mayor sufrimiento, ahora vivir dos veces me dejaria tranquilo porque despues del final de mi vida podria volver a encontrarme contigo y vivir todos los momentos bellos una vez mas y eso seria un sueño volviendose realidad

(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]