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3 years ago

12

Find the area of a rectangle that has a length of 2x+3 and width of x some one painted an area inside the rectangle that has a l

ength and a width of x-2 find the area that was not painted

1 answer:

sashaice [31]3 years ago

8 0

this is the equation and the answer is: 2x+3=5-2=3

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Match the proof part 1:

konstantin123 [22]

Answer:

A, B, C

Step-by-step explanation:

Step 1: "AB ≅ DE, AC ≅ DF, and ∠A ≅ ∠D"

A. Given.

This is the information that was given in the problem statement.

Step 2: "ΔABC ≅ ΔDEF"

B. Side-Angle-Side Postulate (SAS)

The SAS postulate says that if two triangles have a pair of congruent angles between two pairs of congruent sides, then the triangles must be congruent. From the previous step, we can conclude the triangles are congruent.

Step 3: "∠C ≅ ∠F"

C. Corresponding parts of congruent triangles are congruent (CPCTC)

In Step 2, we established the triangles are congruent. So now we can conclude that the corresponding angles are congruent.

3 0

4 years ago

Whoever is looking at this, please help me. i'll give you 19 points!

Lapatulllka [165]

c² = a² + b² - 2 a b cos C

2 a b cos C = a² + b² - c²

cos C = (a² + b² - c²) / 2ab

C = arccos( (a² + b² - c²) / 2ab )

C = arccos( (17² + 8² - 16²) / (2 (17)(8) ) )

C ≈ 69.11°

Answer: 69.1

6 0

4 years ago

Please find the missing side of the triangle and round the answer to the nearest tenth. Thanks.

SashulF [63]

Answer:

39.6

Step-by-step explanation:

Given in the right angled triangle above are:

Ѳ = 49°,

Adjacent length = 26

Hypotenuse length = x

To find x in the right angled triangle given above, apply the trigonometric formula, cos Ѳ = adjacent length/hypotenuse length

Thus,

$cos(49) = \frac{26}{x}$

Multiply both sides by x

$cos(49)*x = \frac{26}{x}*x$

$cos(49)*x = 26$

$0.6561*x = 26$

Divide both sides by 0.6561 to find x

$\frac{0.6561*x}{0.6561} = \frac{26}{0.6561}$

$x = \frac{26}{0.6561}$

$x = \frac{26}{0.6561}$

$x = 39.63$

x = 39.6 (to nearest tenth)

8 0

3 years ago

A men’s softball league is experimenting with a yellow baseball that is easier to see during night games. One way to judge the e

vodomira [7]

Answer:

Yes. There is enough evidence to support the claim that there are fewer errors on average when the yellow ball is used.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>The sample data is:</em>

Yellow 5 2 6 7 2 5 3 8 4 9

White 7 6 8 5 9 11 8 3 6 10

This is a hypothesis test for the difference between populations means.

The claim is that there are fewer errors on average when the yellow ball is used.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2< 0$

The significance level is α=0.05.

The sample 1 (yellow ball errors), of size n1=10 has a mean of 5.1 and a standard deviation of 2.42.

The sample 2 (white balls errors), of size n2=10 has a mean of 7.3 and a standard deviation of 2.41.

The difference between sample means is Md=-2.2.

$M_d=M_1-M_2=5.1-7.3=-2.2$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2+\sigma_2^2}{n}}=\sqrt{\dfrac{2.42^2+2.41^2}{10}}\\\\\\s_{M_d}=\sqrt{\dfrac{11.665}{10}}=\sqrt{1.166}=1.08$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-2.2-0}{1.08}=\dfrac{-2.2}{1.08}=-2.037$

The degrees of freedom for this test are:

$df=n_1+n_2-1=10+10-2=18$

This test is a left-tailed test, with 18 degrees of freedom and t=-2.037, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t$

As the P-value (0.028) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that there are fewer errors on average when the yellow ball is used.

3 0

4 years ago

lisa has 1/3 of an hour to read and watch television. she wants to split her time equally between two activities. how much time

Finger [1]

60 min = one hour

1/3 *60 =20 mins

= 20/2
= 10 min

Lisa has ten minutes

5 0

3 years ago