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3 years ago

Identify whether the following equation has a unique solution, no solution, or infinitely many solutions.

Mathematics

1 answer:

lidiya [134]3 years ago

3 0

Answer:

No solution

Step-by-step explanation:

3+6=9

3-4=-1

9 doesn't equal -1 therefore no solution.

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Find the value of x. please helpp

SIZIF [17.4K]

Answer:

Step-by-step explanation:

given 2 secants drawn from an external point to the circle , then

the product of the measures of one secant's external part and that entire secant is equal to the product of the other secant's external part and that entire secant, that is

9(9 + 2 + 3x) = 10(10 + 2x + 2)

9(11 + 3x) = 10(12 + 2x) ← distribute parenthesis on both sides

99 + 27x = 120 + 20x ( subtract 20x from both sides )

99 + 7x = 120 ( subtract 99 from both sides )

7x = 21 ( divide both sides by 7 )

x = 3

4 0

2 years ago

Read 2 more answers

Find the derivative of the following functions: a. f(x) = (x^3 + 5)^1/4 - 15e^x^3 b. f(x) = (x - 3)^2 (x - 5)/(x - 4)^2(x^2 + 3)

Darya [45]

Answer:

Step-by-step explanation:

Given function is

(a)F(x)= $\left ( x^{3}+5\right )^{0.25}-15e^{x^{3}}$

$F^{'}\left ( x\right )$ = $0.25\left ( x^{3}+5\right )^{-0.75}\frac{\mathrm{d} x^{3}}{\mathrm{d} x}-15e^{x^{3}}\frac{\mathrm{d} x^{3}}{\mathrm{d} x}$

$F^{'}\left ( x\right )=0.25\left ( x^{3}+5\right )^{-0.75}\times 3x^{2}-15e^{x^{3}}\times \left ( 3x^{2}\right )$

(b)F(x)= $\frac{\left ( x-3\right )^2\left ( x-5\right )}{\left ( x-4\right )^2\left ( x^{2}+3\right )^5}$

$F^{'}\left ( x\right )$ = $\frac{\left [ 2\left ( x-3\right )\right \left ( x-5\right )+\left ( x-3\right )^2]\left [ \left ( x-4\right )^2\left ( x^2+3\right )^5\right ]-\left [ 2\left ( x-4\right )^{3}\left ( x^2+3\right )^5+5\left ( x^2+3\right )^4\left ( 2x\right )\left ( x-4\right )^2\right ]\left [ \left ( x-3\right )^2\left ( x-5\right )\right ]}{\left [\left ( x-4\right )^2\left ( x^2+3\right )^5\right ]^2}$

6 0

4 years ago

Can someone answer this please

strojnjashka [21]

the answer is 160 because 8 times 5 equals 40 times it by 4 equals 160

8 0

4 years ago

Simplify: [5×(25)^n+1 - 25 × (5)^2n]/[5×(5)^2n+3 - (25)^n+1]

blagie [28]

$\green{\large\underline{\sf{Solution-}}}$

<u>Given expression is </u>

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 4} - {5}^{2n + 2} }$

can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

<u>Hence, </u>

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

3 0

3 years ago

The coordinates of an ordered pair have opposite signs. In which quadrant(s) must the ordered pair lie?

Virty [35]

If they have opposite signs the point can be in either the second or fourth quadrants.

-x and +y is in the second quadrant

+x and -y is in the fourth quadrant.

3 0

4 years ago

Read 2 more answers