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2 years ago

What is the y-intercept of the line shown?

Mathematics

2 answers:

rewona [7]2 years ago

5 0

Answer:

Step-by-step explanation:

The y-intercept of the line shown is 4.

Hope this helps :)

dexar [7]2 years ago

3 0

Answer: 4

Step-by-step explanation:

It crosses on the y axis at the point (0,4)

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Find a point on the curve x^3+y^3=11xy other than the origin at which the tangent line is horizontal.

attashe74 [19]

Compute the derivative dy/dx using the power, product, and chain rules. Given

x³ + y³ = 11xy

differentiate both sides with respect to x to get

3x² + 3y² dy/dx = 11y + 11x dy/dx

Solve for dy/dx :

(3y² - 11x) dy/dx = 11y - 3x²

dy/dx = (11y - 3x²)/(3y² - 11x)

The tangent line to the curve is horizontal when the slope dy/dx = 0; this happens when

11y - 3x² = 0

y = 3/11 x²

(provided that 3y² - 11x ≠ 0)

Substitute y into into the original equation:

x³ + (3/11 x²)³ = 11x (3/11 x²)

x³ + (3/11)³ x⁶ = 3x³

(3/11)³ x⁶ - 2x³ = 0

x³ ((3/11)³ x³ - 2) = 0

One (actually three) of the solutions is x = 0, which corresponds to the origin (0,0). This leaves us with

(3/11)³ x³ - 2 = 0

(3/11 x)³ - 2 = 0

(3/11 x)³ = 2

3/11 x = ³√2

x = (11•³√2)/3

Solving for y gives

y = 3/11 x²

y = 3/11 ((11•³√2)/3)²

y = (11•³√4)/3

So the only other point where the tangent line is horizontal is ((11•³√2)/3, (11•³√4)/3).

3 0

2 years ago

H(x) = x^2 + 1 k(x) = x – 2<br> (h + k)(2)=?

likoan [24]

First we will compute the h+k and then multiply the result by 2.
To add polynomials, we add terms whose variables are alike, for example:
we add the coefficients of x^2 together, the coefficients of x together and so on.

Therefore:
h + k = x^2 + 1 + x - 2 = x^2+x-1

Now, we will multiply this answer by 2 to get the final answer:
2(h+k) = 2(x^2+x-1) = 2x^2 + 2x -2

6 0

4 years ago

Simplify the expression

ollegr [7]

Answer:

Step-by-step explanation:

$\dfrac{x^2-8x+12}{x^2-7x+10}\\\\x^2-7x+10\ne0\ \iff\ x=\frac{7\pm\sqrt{49-40}}{2}\ne0\ \iff\ x\ne5\ \wedge\ x\ne2\\\\\\\dfrac{x^2-8x+12}{x^2-7x+10}=\dfrac{x^2-2x-6x+12}{x^2-2x-5x+10}=\dfrac{x(x-2)-6(x-2)}{x(x-2)-5(x-2)}=\\\\\\ =\dfrac{(x-2)(x-6)}{(x-2)(x-5)}=\dfrac{x-6}{x-5}\\\\\\f(x)=x-6\\\\g(x)=x-5$

4 0

3 years ago

Alexander deposited money into his retirement account that is compounded annually at an interest rate of 7%.

anzhelika [568]

Tow rates are equivalent if tow initial investments over a the same time, produce the same final value using different interest rates.

For the annually rate we have that:
$V_{0} =(1+ i_{a} ) ^{1}$
Where
$V_{0}$ = initial investment.
$i_{a}$ = annually interest rate in decimal form.
And the exponent (1) represents the full year.

For the quarterly interest rate we have that:
$V_{0} =(1+ i_{q} ) ^{4}$
Where
$V_{0}$ = initial investment.
$i_{q}$ = quarterly interest rate in decimal form.
And the exponent (4) the 4 quarters in the full year.

Since the rates are equivalent if tow initial investments over a the same time, produce the same final value, then
$(1+ i_{a} )=(1+ i_{q} ) ^{4}$
Notice that we are not using the initial investment $V_{0}$ since they are the same.

The first thin we are going to to calculate the equivalent quarterly rate of the 7% annually rate is converting 7% to decimal form
7%/100 = 0.07
Now, we can replace the value in our equation to get:
$(1+0.07)=(1+ i_{q} ) ^{4}$
$1.07=(1+ i_{q} ) ^{4}$
$\sqrt[4]{1.07} =1+ i_{q}$
$i_{q} = \sqrt[4]{1.07} -1$
$i_{q} =0.017$
Finally, we multiply the quarterly interest rate in decimal form by 100% to get:
(0.017)(100%) = 1.7%
We can conclude that Alexander is wrong, the equivalent quarterly rate of an annually rate of 7% is 1.7% and not 2%.

6 0

3 years ago

What is the volume of the square pyramid with base edges 12cm height 10cm

Alina [70]

V=480cm³ . 10 and 12

4 0

3 years ago