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3 years ago

8

Ivory created a scaled copy of triangle C using a scale factor of 4.What is the height of Ivory's copy?The area of the triangle

they gave me is 4.5 units

1 answer:

SOVA2 [1]3 years ago

6 0

Answer:
Height of the Ivory's copy = 12 units
Hope this helps

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1. What is a residual?

sladkih [1.3K]

Answer:

1. what is a residual?

A. A residual is a value of y -y, which is the difference between an observed value of y and a predicted value of y.

2. The regression line has the property that the_sum of squares_of the residuals is the minimum possible sum.

Step-by-step explanation:

1. What is a residual?

A. A residual is a value of y -y, which is the difference between an observed value of y and a predicted value of y.

2. In what sense is the regression line the straight line that "best" fits the points in a scatterplot?

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3 years ago

-25^-1/2 with explanation please

slavikrds [6]

Answer:

The simplified value of $-25^{-\frac{1}{2}}$ is $-\frac{1}{5}$ .

Step-by-step explanation:

As the given expression is

$-25^{-\frac{1}{2}}$

Lets simplify this expression step by step

As

$-25^{-\frac{1}{2}}....[A]$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

$25^{-\frac{1}{2}}=\frac{1}{25^{\frac{1}{2}}}$

So, lets plug $25^{-\frac{1}{2}}=\frac{1}{25^{\frac{1}{2}}}$ in equation [A] i.e. $-25^{-\frac{1}{2}}....[A]$

$-25^{-\frac{1}{2}}....[A]$

$=-\frac{1}{25^{\frac{1}{2}}}$ ∵ $25^{-\frac{1}{2}}=\frac{1}{25^{\frac{1}{2}}}$

$=-\frac{1}{5}$ ∵ $25^{\frac{1}{2}}=5$

Therefore, the simplified value of $-25^{-\frac{1}{2}}$ is $-\frac{1}{5}$ .

Keywords: simplification, exponent rule

Learn more about expression simplification from brainly.com/question/13848834

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3 years ago

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jolli1 [7]

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3 years ago

Is 1/10 rational or irrational

dlinn [17]

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3 years ago

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Determine the longest interval in which the given initialvalue problemis certainto have a unique twicedifferentiable solution. D

AlekseyPX

Answer:

t_o = 3, so solution exists on (0,4).

Step-by-step explanation:

Use Theorem

Divide equation with t(t — 4).

y''+[3/(t-4)]*y'+ [4/t(t-4)]*y=2/t(t-4)

p(t)=3/t-4—> continuous on (-∞, 4) and (4,∞)

q(t) = 4/t(t-4) —> continuous on (-∞,0), (0,4) and (4, ∞)

g(t) = 2/t(t-4)—> continuous on (-∞, 0), (0,4) and (4,∞)

t_o = 3, so solution exists on (0,4).

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3 years ago