All categories

2 years ago

How do u do this please

Mathematics

1 answer:

blagie [28]2 years ago

4 0

Answer: In number 1 the answer is 2. Number 2 the answer is 7. Number 3 you read the problem as 4 x 4 x 4 or 4 to the 3rd. For number 4 it's 2x2x2x2x2x2x2 or 2 to the 7th power. Number 5 is right. For number 6 B. is 5. C is 6

Hope this helps

Step-by-step explanation:

You might be interested in

Help. help. help.<br>no link pls.

kondaur [170]

Answer:

x + 2, x ≠ -2

Step-by-step explanation:

Note that F(x) = x^2 + 4x + 4 factors into (x + 2)^2, and that we are dividing this by g(x) = x + 2. The quotient is x + 2, which in this problem has been named h(x).

The quotient (f/g)(x) is therefore x + 2 EXCEPT that this is not defined at x = -2, because the denominator, x + 2, will be zero there, undefined.

8 0

2 years ago

Show work please!!! Solve the equation by completing the square. If necessary, round to the nearest hundredth.

weeeeeb [17]

Answer:

1. B 2. C

Step-by-step explanation:

1. x^2-18x+(18/2)^2=19+(18/2)^2

(x-18/2)^2=100

(x-9)^2=100

x1= -1 x2= 19

2. x^2=81

x=+-9

x1=-9 x2=9

5 0

2 years ago

Henry has a bag of marbles. six of them are striped. The other half of the marbles are not striped. How many marbles does Henry

Inessa [10]

Answer:

Step-by-step explanation:

6 are striped which is half of them, the other half must be 6 then

6+6=12

5 0

3 years ago

Read 2 more answers

A school is planning to construct two rectangular play areas in the playground. The length of play area A must be 1 foot longer

Anna11 [10]

Answer:

А.The system has two solutions, but only one is viable because the other results in a negative width.

Step-by-step explanation:

Given

Let:

$L_A \to$ length of play area A

$W_A \to$ width of play area A

$L_B \to$ length of play area B

$W_B \to$ width of play area B

$x \to$ Area of A

$y \to$ Area of B

From the question, we have the following:

$L_A = 1 + 4W_A$

$W_B = 2 + W_A$

$L_B = 2 + 3W_B$

$x = y$

The area of A is:

$x = L_A * W_A$

This gives:

$x = (1 + 4W_A) * W_A$

Open bracket

$x = W_A + 4W_A^2$

The area of B is:

$y = L_B * W_B$

$y = (2 + 3W_B) * ( 2 + W_A)$

Substitute: $W_B = 2 + W_A$

$y = (2 + 3(2 + W_A)) * ( 2 + W_A)$

Open brackets

$y = (2 + 6 + 3W_A) * ( 2 + W_A)$

$y = (8 + 3W_A) * ( 2 + W_A)$

Expand

$y = 16 + 8W_A + 6W_A + 3W_A^2$

$y = 16 + 14W_A + 3W_A^2$

We have that:

$x = y$

This gives:

$W_A + 4W_A^2 = 16 + 14W_A + 3W_A^2$

Collect like terms

$4W_A^2 - 3W_A^2 + W_A -14W_A - 16 =0$

$W_A^2 -13W_A - 16 =0$

Using quadratic calculator, we have:

$W_A = -14.1$ or $W = 1.13$ --- approximated

But the width can not be negative; So:

$W = 1.19$

7 0

2 years ago

P(x) = x + 1x² – 34x + 343<br> d(x)= x + 9

Feliz [49]

Answer:

$x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343$

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > $\frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1$ . Go ahead and simplify.

$x=\frac{9}{d-1};\quad \:d\ne \:1$ .

Now, $\mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}$ .

$P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343$ .

Group the like terms... $1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343$ .

$\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1}$ > $1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343$ .

Now for $1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}$ .

$\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}$ .

Now for $33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}$ .

Thus we then get $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343$ .

Now we want to combine fractions. $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}$ .

$\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2$

$\mathrm{For}\:\frac{297}{d-1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:d-1 > \frac{297}{d-1}=\frac{297\left(d-1\right)}{\left(d-1\right)\left(d-1\right)}=\frac{297\left(d-1\right)}{\left(d-1\right)^2}$

$\frac{9^2}{\left(d-1\right)^2}-\frac{297\left(d-1\right)}{\left(d-1\right)^2} > \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}> \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}$ .

Expand $81-297\left(d-1\right) > -297\left(d-1\right) > \mathrm{Apply\:the\:distributive\:law}: \:a\left(b-c\right)=ab-ac$ .

$-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1$ .

$\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}$

Therefore $P=\frac{-297d+378}{\left(d-1\right)^2}+343$ .

Hope this helps!

5 0

3 years ago