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2 years ago

Find the total surface area of the rectangular pyramid.

Mathematics

2 answers:

pantera1 [17]2 years ago

5 0

Answer:

I can't really understand the wording but..

Step-by-step explanation

The formula for surface area of a rectangular prism is A=2(wl+hl+hw)

so yeah enjoy freedom for the rest of the day

VashaNatasha [74]2 years ago

4 0

12 because 12 is never wrong and 12 is an amazing number

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Please help. will give brainliest

Temka [501]

Answer:

84 sqmm

Step-by-step explanation:

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8 0

2 years ago

Read 2 more answers

PLEASE HELP ASAP!!!!!

MakcuM [25]

Answer:

It's significant because it's near the origin

6 0

2 years ago

Fill in the missing equation

Valentin [98]

Start with y=mx+b

B is your y intercept
m is your slope

The points you have are (-1,4) and (1,-2)
To find the slow you Do Y2-Y1/X2-X1
(-2-4)/ (-1-1)
-6/-2
your m is 3

Now you go back to the equation of a Line y=mx+b
to find b (your y intercept) take on of the two points, and substitute it's y value in the equation Let's take (1,-2)

Y=mx+b
-2=3(1)+b
-2=3+b
b=-5

you now know that
b=-5
m=3x

The equation is
y=3x-5

7 0

3 years ago

Solve 17x - 2 = 7x + 8

romanna [79]

17x-2=7x+8

We simplify the equation to the form, which is simple to understand
17x-2=7x+8

We move all terms containing x to the left and all other terms to the right.
+17x-7x=+8+2

We simplify left and right side of the equation.
+10x=+10

We divide both sides of the equation by 10 to get x.
x=1

6 0

2 years ago

Read 2 more answers

For what values of x does the curve y^2x^3-15x^2=4 have horizontal tangent lines

enyata [817]

We have the following curve:

$y^2x^3-15x^2=4$

To find the tangent lines to the curve we need to use the concept of derivative, but we can’t solve this problem for $y$ , thus, let's apply implicit differentiation, so:

$\frac{d}{dx}(y^2x^3-15x^2=4) \\ \\
\frac{d}{dx}(y^2x^3)-\frac{d}{dx}(15x^2)=\frac{d}{dx}(4) \\ \\
(y^{2})'x^3+(x^3)'y^2-15(x^2)'=0 \\ \\ 2yy'x^3+3x^2y^2-30x=0 \\ \\
y'=\frac{dy}{dx}=\frac{30x-3x^2y^2}{2x^3y}=\frac{x(30-3xy^2)}{2x^3y} \\ \\
y'=\frac{30-3xy^2}{2x^2y}$

Therefore the horizontal lines occurs when $y'=0$ , then:

$\frac{30-3xy^2}{2x^2y}=0 \\ \\ That \ is, \
when: \\ \\ 30-3xy^2=0 \\ \\ \therefore xy^2=10 \rightarrow y^2=\frac{10}{x}$

If we substitute this in the original equation we have:

$\frac{10}{x}x^3-15x^2=4 \\ \\ \therefore
10x^2-15x^2=4 \\ \\ \therefore x^2=-\frac{4}{5}$

This is an absurd result because it is impossible for a squared number to get a negative number. So the conclusion is that there is no any value of $x$ in which the curve has horizontal tangent lines.

8 0

2 years ago