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3 years ago

9

7.

1 answer:

anzhelika [568]3 years ago

8 0

The answer is D as said above

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Uhh pls help I got 3 mins left

Ostrovityanka [42]

The side is -15 and the top is 56

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3 years ago

snow_tiger [21]

Answer:

Step-by-step explanation:

the answer is D. 6 ≤ x < ∞

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3 years ago

Giving brainliest, 50 points (answer both questions with solutions)

Helga [31]

Answer:

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3 years ago

Read 2 more answers

A man sold a fan for Rs. 400. Find the cost price if he incurred a loss of 15%.

statuscvo [17]

Answer:

Cost price of the fan is $\text {Rs. 470.59}$ .

Step-by-step explanation:

Given: A man sold a fan for $\text{Rs.}\:400$ . He incurred a loss of $15\%$ .

To find: The cost price of the fan

Solution:

We have,

Selling price, S.P., of the fan $= \text{Rs.} 400$

Loss $=15\%$

We know that $\text {Loss\%}=\frac{\text{C.P.-S.P.}}{\text{C.P.}}\times100\%$

Now let the C.P. of the fan be $\text{Rs.} \:x$

So, $15=(\frac{x-400}{x})\times 100$

$\implies 15x=100x-40000$

$\implies 100x-15x=40000$

$\implies 85x=40000$

$\implies x=\frac{40000}{85}$

$\implies x= 470.59$

Hence, the cost price of the fan is $\text {Rs. 470.59}$ .

7 0

3 years ago

It is possible to get 2 solutions when the system of equations is: (check all that apply)

SVETLANKA909090 [29]

Answer:

A system of the equation of a circle and a linear equation

A system of the equation of a parabola and a linear equation

Step-by-step explanation:

Let us verify our answer

A system of the equation of a circle and a linear equation

Let an equation of a circle as $x^2+ y^2 = 1$ ..........(1)

Let a liner equation Y = x ............(2)

substitute (2) in (1)

$x^2 + x^2 = 1\\2x^2 = 1\\$

$x^2 = \frac{1}{\sqrt{2} } \\x = +\frac{1}{\sqrt{2} } , -\frac{1}{\sqrt{2} }$ so Y = $+\frac{1}{\sqrt{2} } , -\frac{1}{\sqrt{2} }$

so the two solution are ( $(\frac{1}{\sqrt{2} } ,\frac{1}{\sqrt{2} }) (-\frac{1}{\sqrt{2} }, -\frac{1}{\sqrt{2} }$ )

A system of the equation of a parabola and a linear equation

Let equation of Parabola be $y^2 = x$

and linear equation y = x

substitute

$x^2 = x\\x^2 - x= 0\\x(x-1) = \\x = 0 , 1$

Y = 0,1

so the two solutions will be (0,0) and (1,1)

3 0

3 years ago