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3 years ago

15

a triangle has coordinates a(-1,-2) b(-4,-2) and c(-4,-5) what are the coordinates of a' the image of point a under a dilation w

ith a scale factor of 3?

1 answer:

quester [9]3 years ago

8 0

9514 1404 393

Answer:

(-3, -6)

Step-by-step explanation:

The scale factor multiplies each coordinate.

A' = 3A = 3(-1, -2) = (-3, -6)

The image of A is ...

A'(-3, -6)

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Consider testing Upper H 0 : mu equals 20H0: μ=20 against Upper H Subscript a Baseline : mu less than 20Ha: μ<20 where muμ is

kolezko [41]

Answer:

Test: $\mu$ < 20 with normal distⁿ

Hypothesis test:

H0: $\mu \ge$ 20 (Null Hypothesis, H₀ : $\mu$ = 20 is also correct)

H1: $\mu$ < 20 (Alternative Hypothesis, also called H₁)

This is lower tailed test.

Since sample is large, sample standard deviation can be taken as an approximation of population standard deviation.

x = 19.3

$\sigma$ = 11.1

n = 444

significance level, $\alpha$ = 0.05 (If no value is given, we take level of 0.05)

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3 years ago

Given the equation 8x-6+3x=60. Sasha claims the next step will result 5x=66. Dante claims it will be 11x=66 Who is right and why

IrinaVladis [17]

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Question 20 not yet answered marked out of 1.00 not flaggedflag question question text luna created a trash can in the shape of

Alex Ar [27]

Consider the triangle with sides a=1.6, b=2.3 and c=1.2 feet. You can use cosine theorem twice to find two unknown angles:

1.

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Then you can conclude that ∠B is obtuse (because $\cos \angle B$ ) and m∠B=110°.

2.

$c^2=a^2+b^2-2ab\cos \angle C,\\ (1.2)^2=(1.6)^2+(2.3)^2-2\cdot 1.6\cdot 2.3\cos \angle C,\\ 1.44=2.56+5.29-7.36\cos \angle C,\\ 7.36\cos \angle C=7.85-1.44=6.41,\\\cos \angle C=\dfrac{6.41}{7.36} =0.871$ .

Then you can conclude that ∠C is acute (because $\cos \angle C>0$ ) and m∠C=29°.

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den301095 [7]

Answer:

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