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3 years ago

7

Can someone answer this i need help

2 answers:

NISA [10]3 years ago

4 0

The answer is 6x^6 - 10x^3 + 8x^2

Oxana [17]3 years ago

3 0

All that is needed to do is to simplify the expression.

The answer to your question is:

$6 x^{6}$ - $10 x^{3}$ + $8x^{2}$

Hope I helped!

You might be interested in

Please help me 33/2 + 3y/5 = 7y/10 + 15

8090 [49]

We are given with an equation in <em>variable y</em> and we need to solve for <em>y</em> . So , now let's start !!!

We are given with ;

${:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}$

Take LCM on both sides :

${:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}$

<em>Multiplying</em> both sides by <em>10</em> ;

${:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}$

${:\implies \quad \sf 165+6y=7y+150}$

Can be <em>further written</em> as ;

${:\implies \quad \sf 7y+150=165+6y}$

Transposing <em>6y </em>to<em> LHS</em> and <em>150</em> to<em> RHS </em>

${:\implies \quad \sf 7y-6y=165-150}$

${:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}$

8 0

2 years ago

A rectangular box with a square base is to be constructed from material that costs $9 per ft2 for the bottom, $5 per ft2 for the

kramer

Answer:

$18.73ft^3$

Step-by-step explanation:

Let

Side of square base=x

Height of rectangular box=y

Area of square base=Area of top= $(side)^2=x^2$

Area of one side face= $l\times b=xy$

Cost of bottom=$9 per square ft

Cost of top=$5 square ft

Cost of sides=$4 per square ft

Total cost=$204

Volume of rectangular box= $V=lbh=x^2y$

Total cost= $9(x^2)+5x^2+4(4xy)=14x^2+16xy$

$204=14x^2+16xy$

$204-14x^2=16xy$

$y=\frac{204-14x^2}{16x}=\frac{102-7x^2}{8x}$

Substitute the values of y

$V(x)=x^2\times \frac{102-7x^2}{8x}=\frac{1}{8}(102x-7x^3)$

Differentiate w.r.t x

$V'(x)=\frac{1}{8}(102-21x^2)=0$

$V'(x)=0$

$\frac{1}{8}(102-21x^2)=0$

$102-21x^2=0$

$102=21x^2$

$x^2=\frac{102}{21}=4.85$

$x=\sqrt{4.85}=2.2$

It takes positive because side length cannot be negative.

Again differentiate w.r. t x

$V''(x)=\frac{1}{8}(-42x)$

Substitute the value

$V''(2.2)=-\frac{42}{8}(2.2)=-11.55$

Hence, the volume of box is maximum at x=2.2 ft

Substitute the value of x

$y=\frac{102-7(2.2)^2}{8(2.2)}=3.87 ft$

Greatest volume of box= $x^2y=(2.2)^2\times 3.87=18.73 ft^3$

5 0

3 years ago

I KNOW IT IT LONG BUT PLEASE HELP ME PLEASE

S_A_V [24]

See I have had bad days in school Bc of so much homework and writing but thissssss

3 0

3 years ago

BRAINLIEST PLS HELP

Ann [662]

C. 1 1/3
7/9 + 5/9 = 12/9
12/9 = 1 3/9 = 1 1/3

5 0

2 years ago

Read 2 more answers

You flip a coin and then roll a 6-sided number cube (a die).

expeople1 [14]

Answer:

a) No, it does not matter whether you roll the die or flip the coin first, as these two events are <u>independent</u> of each other, which means they do not affect each other.

b) Yes.

Let event 1 be flipping a coin and event 2 be rolling a die.
Let event 1 be rolling a die and event 2 be flipping a coin.

The likelihood that any outcome will occur will not change, as the events are independent.

c) see attached

d) 12 outcomes (H = head, T = tail, numbers represent the value of the die)

H 1 T 1

H 2 T 2

H 3 T 3

H 4 T 4

H 5 T 5

H 6 T 6

e)

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

$\implies \sf P(even)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}$

$\implies \sf P(head)=\dfrac{1}{2}$

$\implies \sf P(even)\:and\:P(head)=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}$

6 0

2 years ago