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3 years ago

1. Giles thinks that -5x²(3x - 4) simplifies to 15x² + 20x². Explain what mistakes he made. What is the

Mathematics

1 answer:

jeka57 [31]3 years ago

6 0

Answer:

The mistake was minus sign before the 15 and putting 15x² instead of 15x³.

The correct answer is:

$-5x^{2} (3x-4)\\ = -15x^{3} + 20x^{2}$

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Jakob simplified the expression as shown.

Flauer [41]

Answer:

He simplified the x coefficients incorrectly

Step-by-step explanation:

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2 years ago

Read 2 more answers

The surface area of a cone is 54 square inches. What is the surface area of a similar cone that is three times as large?

12345 [234]

<u>Given</u>:

Given that the surface area of the cone is 54 square inches.

We need to determine the surface area of the cone that is similar to the cone three times large.

<u>Surface area of the similar cone:</u>

Let us determine the surface area of the similar cone.

The surface area of the similar cone can be determined by multiplying the surface area of the cone by 3. Because it is given that the similar cone is three times large.

Thus, we have;

$SA = 3 \times 54$

$SA=162$

Thus, the surface area of the similar cone is 162 square inches.

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3 years ago

What is the equation for the plane illustrated below?

TiliK225 [7]

Answer:

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

$a\cdot x + b\cdot y + c\cdot z = d$

Where:

$x$ , $y$ , $z$ - Orthogonal inputs.

$a$ , $b$ , $c$ , $d$ - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0), (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

$y = m\cdot x + b$

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - x-Intercept, dimensionless.

If $x_{1} = 2$ , $y_{1} = 0$ , $x_{2} = 0$ and $y_{2} = 2$ , then:

Slope

$m = \frac{2-0}{0-2}$

$m = -1$

x-Intercept

$b = y_{1} - m\cdot x_{1}$

$b = 0 -(-1)\cdot (2)$

$b = 2$

The equation of the line in the xy-plane is $y = -x+2$ or $x + y = 2$ , which is equivalent to $3\cdot x + 3\cdot y = 6$ .

yz-plane (0, 2, 0) and (0, 0, 3)

$z = m\cdot y + b$

$m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}$

Where:

$m$ - Slope, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - y-Intercept, dimensionless.

If $y_{1} = 2$ , $z_{1} = 0$ , $y_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

y-Intercept

$b = z_{1} - m\cdot y_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the yz-plane is $z = -\frac{3}{2}\cdot y+3$ or $3\cdot y + 2\cdot z = 6$ .

xz-plane (2, 0, 0) and (0, 0, 3)

$z = m\cdot x + b$

$m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - z-Intercept, dimensionless.

If $x_{1} = 2$ , $z_{1} = 0$ , $x_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

x-Intercept

$b = z_{1} - m\cdot x_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the xz-plane is $z = -\frac{3}{2}\cdot x+3$ or $3\cdot x + 2\cdot z = 6$

After comparing each equation of the line to the definition of the equation of the plane, the following coefficients are obtained:

$a = 3$ , $b = 3$ , $c = 2$ , $d = 6$

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

8 0

3 years ago

Area of a semicircle is 308cm. Find the perimeter

fgiga [73]

Answer:

72 cm

Step-by-step explanation:

Find the radius using the given information

308=1/2(πr^2)

r^2=616/π

r=14

The perimeter is πr + 2r so plug in 14

π(14) + 2(14)

43.96+28=71.96

3 0

3 years ago

A rectangle has dimensions of 12 cm long and 10 cm wide what is the area of the rectangle

LiRa [457]

The area and the width are known as 60 and 12 respectively. Rearranging that equation above, and plugging the values, one gets the value of the length of the rectangle as 5.

3 0

3 years ago