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zalisa [80]
2 years ago
8

What is b and c in this expression?

le="\sqrt[4]{15^7} }" alt="\sqrt[4]{15^7} }" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
ozzi2 years ago
8 0

The value of b and  c is 4 and 7.

<h3>What is algebra?</h3>

Algebra is the part of mathematics in which letters and other general symbols are used to represent numbers and quantities in formulae and equations.

Algebra is divided into different sub-branches such as elementary algebra, advanced algebra, abstract algebra, linear algebra, and commutative algebra.

Elementary algebra: elementary algebra, branch of mathematics that deals with the general properties of numbers and the relations between them. Algebra is fundamental not only to all further mathematics and statistics but to the natural sciences, computer science, economics, and business.

From the given question,

\sqrt[4]{15^7} This expression can also be written in the form of a^\frac{b}{c}

We will convert \sqrt[4]{15^7} to the required form:

(15^\frac{7c}{4b} )\left \{ {{a=15^\frac{7c}{4b} } \atop {b=b} , c = c} \right.

Hence,

From the above equation we get the value of b and c is 4 and 7.

Learn more about algebra here:brainly.com/question/17260216

#SPJ1

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Answer the following questions about the sphere whose equation is given by x2+y2+z2−8x+4y=−4. 1. Find the radius of the sphere.
Anna [14]

Answer:

we have (a,b,c)=(4,-2,0) and R=4 (radius)

Step-by-step explanation:

since

x²+y²+z²−8x+4y=−4

we have to complete the squares to finish with a equation of the form

(x-a)²+(y-b)²+(z-c)²=R²

that is the equation of a sphere of radius R and centre in (a,b,c)

thus

x²+y²+z²−8x+4y=−4

x²+y²+z²−8x+4y +4 = 0

x²+y²+z²−8x+4y +4 +16-16 =0

(x²−8x + 16) + (y² + 4y + 4 ) + (z²) -16 = 0

(x-4)² + (y+2)² + z² = 16

(x-4)² + (y-(-2))² + (z-0)² = 4²

thus we have a=4 , b= -2 , c= 0 and R=4

8 0
3 years ago
What is 43.2÷16= in partial quotients​
sergejj [24]

43.2\16= 2.7 hope this helps

8 0
3 years ago
Read 2 more answers
Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut
frozen [14]

Answer:

a)\sqrt{144-288t+208t^2} b.) -12knots, 8 knots c) No e)4\sqrt{13}

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved 12\dot t (n.m) and ship B has moved 8\dot t (n.m). We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}

b)

We want to find \frac{ds}{dt} for t=0 and t=1

\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0

Since function f(x)=199-288x+208x^2 is quadratic, concave up and has no real roots, we know that 199-288x+208x^2>0 for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

                                                \lim_{t \to \infty} \frac{ds}{dt}

So, lets check this limit:

\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}

Notice that:

4\sqrt{13}=\sqrt{12^2+5^2}=√(speed of ship A² + speed of ship B²)

5 0
3 years ago
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Amanda [17]
I think it’s dy/dx=2
4 0
3 years ago
A songwriter gets paid monthly at a rate of $150 for each song he completes. Last month he wrote 8 songs and got halfway through
DaniilM [7]

Answer:

The answer would be 1275 if they get paid for half the money on the 9th song

Step-by-step explanation:

3 0
3 years ago
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