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4 years ago

Can you help, please.?

Mathematics

1 answer:

zhenek [66]4 years ago

8 0

Answer:

No 1 inch cubes will fit

28 one-half inch cubes will fit

Step-by-step explanation:

No cubes with a side length of 1 inch will fit because one dimension of the prism is only 1/2 inches, so we can't fit a 1 inch cube in anywhere.

Since one dimension is 1/2 inch, we can line up 1/2 inch cubes along the other 2 dimensions. Let's say that the 1/2 inch dimension is the base, the 2 inch dimension is the height, and the 7/2 inch dimension is the length.

We can stack the cubes 4 high, (2 inches of height with 1/2 inch being the cube height, there are 4 halves in 2).

We can place 7 cubes along the length (7/2 inches = 3.5 inches. There are 7 halves in 3.5).

So we can have 4(7) = 28 one half inch cubes can fit in to the prism

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Given: D is the midpoint of AB; E is the midpoint of AC.

Thepotemich [5.8K]

Answer:

The lines DE and BC have the same slope, therefore the two lines, DE and BC, are parallel

Step-by-step explanation:

To prove that DE is parallel to BC, we have;

The slope, m of the lines DE and BC are found from the following equation;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where;

(x₁, y₁) and (x₂, y₂) = (b, c) and (a + b + c) for DE, we have;

$Slope, \, m =\dfrac{c - c}{a + b-b} = 0$

Where we have (x₁, y₁) and (x₂, y₂) = (0, 0) and (2a, 0) for BC, we have;

$Slope, \, m =\dfrac{0 - 0}{2a-0} = 0$

Therefore, because the lines DE and BC have the same slope, the two lines DE and BC are parallel.

5 0

3 years ago

Read 2 more answers

The length of a new rectangular playing field is 3 yards longer than double the width. If the perimeter of the rectangular playi

Romashka-Z-Leto [24]

Answer:

121 x 59 (Dimensions)

Step-by-step explanation:

6x + 6 = 360

Take away 6 from both sides

6x = 354

Divide both sides by 6

x = 59

The length is 2x + 3 because it says 3 more than double the width (lenght is x). You have 2 widths and 2 lengths. The two widths are both x (2x). The two lengths are both 2x + 3 (4x + 6). 2x + 4x + 6 = 6x + 6.

Now input 59 for x

2(59) + 3 = 121

x = 59

Check Your Answer

121 is just one length and 59 is just one width, so you have to multiply both sides by 2 because you have 2 widths and 2 lengths.

121 x 2 = 242

59 x 2 = 118

242 + 118 = 360

8 0

3 years ago

Solve the following using Substitution method 2x – 5y = -13 3x + 4y = 15

Digiron [165]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{array} { l } { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$2x-5y=-13, \: 3x+4y=15$

Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

$2x-5y=-13$

Add 5y to both sides of the equation.

$2x=5y-13$

Divide both sides by 2.

$x=\frac{1}{2}\left(5y-13\right) \\$

Multiply $\frac{1}{2}\\$ times 5y - 13.

$x=\frac{5}{2}y-\frac{13}{2} \\$

Substitute $\frac{5y-13}{2}\\$ for x in the other equation, 3x + 4y = 15.

$3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15 \\$

Multiply 3 times $\frac{5y-13}{2}\\$ .

$\frac{15}{2}y-\frac{39}{2}+4y=15 \\$

Add $\frac{15y}{2} \\$ to 4y.

$\frac{23}{2}y-\frac{39}{2}=15 \\$

Add $\frac{39}{2}\\$ to both sides of the equation.

$\frac{23}{2}y=\frac{69}{2} \\$

Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

$\large \underline{ \underline{ \sf \: y=3 }}$

Substitute 3 for y in $x=\frac{5}{2}y-\frac{13}{2}\\$ . Because the resulting equation contains only one variable, you can solve for x directly.

$x=\frac{5}{2}\times 3-\frac{13}{2} \\$

Multiply 5/2 times 3.

$x=\frac{15-13}{2} \\$

Add $-\frac{13}{2}\\$ to $\frac{15}{2}\\$ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

$\large\underline{ \underline{ \sf \: x=1 }}$

The system is now solved. The value of x & y will be 1 & 3 respectively.

$\huge\boxed{ \boxed{\bf \: x=1, \: y=3 }}$

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3 years ago

Dominic earns $285 per week plus an 8% commission rate on all his sales.If Dominic sells $4,213 worth of merchandise in one week

irakobra [83]

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3 years ago

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Which of the following is a polynomial with roots 3, 5i, and −5i?

jek_recluse [69]

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3 years ago