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2 years ago

MATHH!!! find the surface area of the prism. pleaseee answer correctly and tell me how you got your answer.

Mathematics

1 answer:

timofeeve [1]2 years ago

7 0

Answer:232 inch^2

Step-by-step explanation:to solve, add the surface area of each side. Remember, there are 3 pairs of sides and the pairs have equal area. Area =length x length. So we have 2x(8x2)+2x(2x10)+2x(10x8) in square inches.

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Which of the following statements is false? (5 points)

Aliun [14]

Answer:

Step-by-step explanation:

The sum of two irrational numbers is always still irrational.

5√3 + 6√5 is still going to be irrational. You cannot find two such numbers adding to rational.

3 0

3 years ago

Read 2 more answers

X + 2y = –6<br> 3x + 8y = –20

Kay [80]

What do you want to calculate?

CALCULATE IT!
Solve
Graph
Lesson

Problem:
Solve
x
+
2
y
=
−
6
;
3
x
+
8
y
=
−
20
Steps:
I will solve your system by substitution.
(You can also solve this system by elimination.)
x
+
2
y
=
−
6
;
3
x
+
8
y
=
−
20
Step: Solve
x
+
2
y
=
−
6
for x:
x
+
2
y
+
−
2
y
=
−
6
+
−
2
y
(Add -2y to both sides)
x
=
−
2
y
−
6
Step: Substitute
−
2
y
−
6
for
x
in
3
x
+
8
y
=
−
20
:
3
x
+
8
y
=
−
20
3
(
−
2
y
−
6
)
+
8
y
=
−
20
2
y
−
18
=
−
20
(Simplify both sides of the equation)
2
y
−
18
+
18
=
−
20
+
18
(Add 18 to both sides)
2
y
=
−
2
2
y
2
=
−
2
2
(Divide both sides by 2)
y
=
−
1
Step: Substitute
−
1
for
y
in
x
=
−
2
y
−
6
:
x
=
−
2
y
−
6
x
=
(
−
2
)
(
−
1
)
−
6
x
=
−
4
(Simplify both sides of the equation)
Answer:
x
=
−
4
and
y
=
−
1

3 0

4 years ago

solve for x. if anyone could explain how exactly I solve this question, it would be greatly appreciated!!! thank you

Gennadij [26K]

Answer:x=29

Step-by-step explanation:

You want to add all the like terms together so in this case 2x and x

29,30,20,8

3x=87

Divide 3 by both sides

X=29

5 0

3 years ago

The condition_______?proves that ∆ABC and ∆EFG are congruent by the SAS criterion.

snow_lady [41]

Answer:

(1) D.Angle C is congruent to to Angle F. (2) C. SSS. (3) C. cannot be congruent to.

Step-by-step explanation:

From the given figure it is noticed that

$AC=EG$

$CB=GF$

According to SAS postulate, if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then both triangles are congruent.

The included angles of congruent sides are angle C and angle G.

So, condition "Angle C is congruent to to Angle F" will prove that the ∆ABC and ∆EFG are congruent by the SAS criterion.

If $AB\neq EF$

According to SSS postulate, if all three sides in one triangle are congruent to the corresponding sides in the other.

Since two corresponding sides are congruent but third sides of triangles are not congruent, therefore SSS criterion for congruence is violated.

Since two corresponding sides are congruent but third sides of triangles are not congruent, therefore the included angle of congruent sides are different.

$\angle C\neq \angle G$

Therefore angle C and angle F cannot be congruent to each other.

4 0

3 years ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago