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3 years ago

7

Part B: If a cross section of the rectangular pyramid is cut perpendicular to the base, passing through the top vertex, what wou

ld be the shape of the resulting cross section? Explain your answer.

1 answer:

Katyanochek1 [597]3 years ago

6 0

Answer:

A triangle.

Step-by-step explanation:

The plane would pass through opposite sloping faces of the pyramid and through the base; the shape formed is a triangle.

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A box contains 2 red marbles , 3 green marbles , and 3 blue marbles

krek1111 [17]

Answer:

the box contains a total of 8 marbles .

although if you understand pixels then you would know that could stand for RGB so your color would be a very dark greenish brown

Step-by-step explanation:

no question?

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3 years ago

Tomas bought 80 tickets for rides at an amusement park. Each ride costs 5 tickets, and Tomas has been on x rides so far. Which e

Alex Ar [27]

Answer:

Option A, C and D are correct choices.

Step-by-step explanation:

Let x be the number of rides that Tomas has been on so far. Each ride costs 5 tickets. This means that cost of x rides will be 5x.

We are told that Tomas bought 80 tickets for rides at an amusement park.

To find the number of of tickets that Tomas has left we will subtract cost of x rides from total number of tickets that Thomas bought.

We can represent this information in an expression as: $80-5x$

Now let us see which of our given choices in equivalent to our expression.

A. $80-5x$

Upon looking at option A we can see that it is same as our expression, therefore, option A is the correct choice.

B. $80+5x$

We can see that in this expression cost of x rides in being added instead of subtraction, therefore, option B is not a correct choice.

C. $5(16-x)$

Upon distributing 5 we will get,

$80-5x$

Now this expression is same same as our expression, therefore, option C is the correct choice.

D. $-5x+80$

This expression is also same as our expression as we can rearrange the terms in this expression as: $80-5x$ , therefore, option D is a correct choice as well.

E. $5(16+x)$

Upon distributing 5 we will get,

$80+5x$

We can see that in this expression cost of x rides in being added instead of subtraction, therefore, option E is not a correct choice.

6 0

3 years ago

jim is planning a party. he spends $150 on a room for the party. he also spends $16.50 per person for food. jim finds the total

meriva

$577.50is the amount he paid.

8 0

3 years ago

Click and drag like terms onto each other to simplify fully. 1+x^2-3+2y+7x^2+3+3y

Rashid [163]

Answer:

8x^2 + 5y + 1

Step-by-step explanation:

1 + x^2 - 3 + 2y + 7x^2 + 3 + 3y

group like terms

= x^2 + 7x^2 + 2y + 3y + 1 - 3 + 3

add similar items

= 8x^2 + 2y + 3y + 1 - 3 + 3

add similar items

= 8x^2 + 5y + 1 - 3 + 3

= 8x^2 + 5y + 1

6 0

3 years ago

Read 2 more answers

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

3 years ago