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4 years ago

Which of the following equations is equivalent to b^y=x

Mathematics

1 answer:

zalisa [80]4 years ago

8 0

Step-by-step explanation:

$\text{De}\text{finition of logarithm}:\\\\\log_ab=c\iff a^c=b,\ \text{where}\ a>0\ \wedge\ a\neq1\ \wedge\ b>0\\\\\text{Therefore your answer is}\\\\\large\huge{\boxed{B.\ y=\log_bx}}\\\\\log_bx=y\iff b^y=x$

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if the total angle of a triangle is 180 A- 43 1/2 B-5x C-26.5 and the started value is 20 what is X?

ValentinkaMS [17]

Answer:

x= 20.05 degree

180= 43.25+36.50+5x

180= 79.75+5x

5x=180-79.75

5x=100.25

x=20.25

ANGLE B= 100.25

3 0

3 years ago

You want to get from a point A on the straight shore of the beach to a buoy which is 54 meters out in the water from a point B o

anyanavicka [17]

Answer:

$x =\dfrac{45 \sqrt{6}}{ 2}$

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

$\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}$

In order to estimate the point along the shore, x meters from B, the runner should stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

The differential of V(x) = V'(x) =0

= $\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0$

$-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0$

$\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}$

squaring both sides; we get

$\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}$

$\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}$

By cross multiplying; we get

$49x^2 = 25(54^2+x^2)$

$49x^2 = 25 \times 54^2+ 25x^2$

$49x^2-25x^2 = 25 \times 54^2$

$24x^2 = 25 \times 54^2$

$x^2 = \dfrac{25 \times 54^2}{24}$

$x =\sqrt{ \dfrac{25 \times 54^2}{24}}$

$x =\dfrac{5 \times 54}{\sqrt{24}}$

$x =\dfrac{270}{\sqrt{4 \times 6}}$

$x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}$

$x =\dfrac{45 \sqrt{6}}{ 2}$

8 0

3 years ago

Anwser the following question with explanation

shepuryov [24]

Answer:

you need to devide the 2 numbers well the 2 fractions bc there mixd numbers well bye it was my pluesure to help you

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Can anyone help me ? I just need the numbers

Nimfa-mama [501]

Answer:

(-6, 0) and (0, 1.5)

Step-by-step explanation:

The equation of the line in pint slope form is expressed as;

y-y0= m(x-x0)

m is the slope

(x0, y0) is the point on the line

Given

m = 1/4

(x0, y0) = (6,3)

Substitute into the formula;

y - 3 = 1/4(x-6)

4(y-3) = x - 6

4y - 12 = x-6

4y - x = -6+12

4y - x = 6

x = 4y - 6

To get the points to plot, we will find the x and y-intercept of the resulting expression.

For the x-intercept,

at y = 0

x = 4(0) - 6

x = -6

Hence the x-intercept is at (-6, 0)

For the y-intercept,

at x = 0

0 = 4y - 6

4y = 6

y = 6/4

y = 3/2

y = 1.5

Hence the y-intercept is at (0, 1.5)

Hence the required points to plot to get the required line are (-6, 0) and (0, 1.5)

7 0

3 years ago

What is the area of this trapezoid?

Sonbull [250]

Find the area of the square first
Area = Length x Width
Area = 12 x 9
Area = 108

Find the area of the right triangles
Area = 1/2(Length x Base)
Area = 1/2(9 x 2)
Area = 1/2(18)
Area = 9
Since there is 2 triangles multiply by 2
9 x 2 = 18

Now add
108 + 18 = 126
126 is your area

8 0

3 years ago

Read 2 more answers