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3 years ago

What number is 64% of 120? i don’t need solutions , just answer please and thank you !!! ❤️

Mathematics

1 answer:

Ivenika [448]3 years ago

4 0

<em>Answer:</em>

<em>Step-by-step explanation:</em>

<em>Hi there !</em>

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Find the surface area of the sphere with the given dimension. Leave your answer in terms of pi

AfilCa [17]

A you just divide bro

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2 years ago

Christina goes to market with $50. She buys one papaya for $20 and spends the rest of the money on bananas. Each banana costs $6

maria [59]

$x-\ number\ of\ bananas\\\\
20+6x\leq50\ \ \ \ | subtract\ 20\\\\
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2 years ago

How many times does the equation f(x)=2x^2+9x-5 intersect the x-axis

Komok [63]

Answer:

Step-by-step explanation:

f(x)=2x^2+9x-5

When we are find how many times it intersects the x axis, we are finding the zero's. Set the equation equal to zero

0=2x^2+9x-5

Factor the equation

0 = (2x+1) (x-5)

2*1

1*-5 = -5

2*-5 +1*1 = -9

This checks for the first last and middle terms so we factored correctly

Then using the zero product property

2x+1 = 0 and x-5 =0

2x = -1 x=5

x = -1/2 and x=5

This function crosses the x axis 2 times

8 0

3 years ago

Solution for -2 1/3 divided by 4 2/3

Zina [86]

The answer would be -0.5 hope this helped!

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2 years ago

Read 2 more answers

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*

pickupchik [31]

<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation $x^4-41x^2=-400$ .

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

$x^4-41x^2+400=-400+400$ .

$x^4-41x^2+400=0$ .

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

$x^4-25x^2-16x^2+400=0$

Making it into two groups, we get

$(x^4-25x^2)+(-16x^2+400)=0$

Factoring out GCF of each group, we get

$x^2(x^2-25)-16(x^2-25)=0$

$(x^2-25)(x^2-16) =0$

Factoring out $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$ , we get

$(x^2-25) = x^2-5^2= (x-5)(x+5)$ and

$x^2-16 = x^2-4^2 =(x-4)(x+4)$ .

Therefore,

$(x-5)(x+5)(x-4)(x+4) =0$

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

6 0

2 years ago