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mihalych1998 [28]

2 years ago

6

On a coordinate plane, a parallelogram has points A (negative 3, 4), B (3, 4), C (1, negative 2), and D (negative 5, negative 2)

. If a translation of (x, y) → (x + 6, y – 10) is applied to figure ABCD, what are the coordinates of D'? (–5, –2) (1, –12) (4, –15) (–9, –6)

2 answers:

EleoNora [17]2 years ago

5 0

Answer:

(1,-12)

Step-by-step explanation:

vladimir2022 [97]2 years ago

3 0

Answer:

(1,-12)

Step-by-step explanation:

d starts at (-5,-2) if you add 6 to -5 and minus -2 by 10 you will end up with (1,-12)

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What pages will be favored for the given search? Search terms: Michael OR Jordan A. Pages about Michael Jordan B. Pages about Mi

ad-work [718]

Answer:

A. Pages about Michael Jordan.

Step-by-step explanation:

Michael Jordan, MJ, is a great basketball player, and has achieved one the best records in his career. He is a National Basketball Association (NBA) player with great skills and energy.

From the search item, the two names Michael OR Jordan is for one person, Michael Jordan. The search would combine the two names because it is a well known one and give an output on Michael Jordan. Thus, the pages that would be favored are pages about Michael Jordan.

7 0

2 years ago

Select the correct answer.

lakkis [162]

Answer:

6x + 4y ≤ 50

x + y ≥ 10

Step-by-step explanation:

$6 is the cost of stuffed animals $4 is the cost of toy trucks and the her maximum budget is $50 it would be 6x+4y is less then or equal to 50 and there is AT LEAST 10 people so the amounts which are x and y would be equal to or greater than 10.

8 0

2 years ago

Read 2 more answers

11) The Cost of maintaing a

dezoksy [38]

Answer:

a). Cost of 44 pupils = $14265

b). Least number of pupils = 31

Step-by-step explanation:

The given question is incomplete; here is the complete question.

The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.

(a) Find the cost when there are 44 pupils.

(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?

Let the equation representing the total cost of maintaining a school is,

C = ax + b

Where C = Total cost of maintaining a school

a = Fee per pupil

b = Fixed running cost

x = number of pupils

a). Cost of 50 pupils = $15705

Equation will be,

15705 = 50a + b -------(1)

Cost of 40 pupils = $13305

Equation will be,

13305 = 40a + b --------(2)

By subtracting equation (2) from equation (1),

15705 - 13305 = (50a + b) - (40a + b)

2400 = 10a

a = 240

From equation (1),

b = 3705

Equation representing the total cost will be,

C = 240x + 3705

If x = 44

C = 240(44) + 3705

C = $14265

b). If the fee per pupil 'a' = $360

Let the number of pupils = p

Total fee of 'p' pupils = $360p

Total cost to run the school will be = 3705 + 240p

For the school not to be in the loss,

360p ≥ 3705 + 240p

360p - 240p ≥ 3705

120p ≥ 3705

p ≥ $\frac{3705}{120}$

p ≥ 30.875

Therefore, to run the school without loss, number pupils should be at least 31.

3 0

2 years ago

Milton spilled some ink on his homework paper. He can't read the coefficient of $x$, but he knows that the equation has two dist

Lera25 [3.4K]

Answer:

$Sum = -81$

Step-by-step explanation:

See the comment for complete question.

Given

$c = 36$ ----- Constant

No coefficient of x^2

Required:

Determine the sum of all distinct positive integers of the coefficient of x

Reading through the complete question, we can see that the question has 3 terms which are:

x^2 ---- with no coefficient

x ---- with an unknown coefficient

36 ---- constant

So, the equation can be represented as:

$x^2 + ax + 36 = 0$

Where a is the unknown coefficient

From the question, we understand that the equation has two negative integer solution. This can be represented as:

$x = -\alpha$ and $x = -\beta$

Using the above roots, the equation can be represented as:

$(x + \alpha)(x + \beta) = 0$

Open brackets

$x^2 + (\alpha + \beta)x + \alpha \beta = 0$

To compare the above equation to $x^2 + ax + 36 = 0$ , we have:

$a = \alpha + \beta$

$\alpha \beta = 36$

Where: $\alpha, \beta$ and $\alpha \ne \beta$

The values of $\alpha$ and $\beta$ that satisfy $\alpha \beta = 36$ are:

$\alpha = -1$ and $\beta = -36$

$\alpha = -2$ and $\beta = -18$

$\alpha = -3$ and $\beta = -12$

$\alpha = -4$ and $\beta = -9$

So, the possible values of a are:

$a = \alpha + \beta$

When $\alpha = -1$ and $\beta = -36$

$a = -1 - 36 = -37$

When $\alpha = -2$ and $\beta = -18$

$a = -2 - 18 = -20$

When $\alpha = -3$ and $\beta = -12$

$a = -3 - 12 = -15$

When $\alpha = -4$ and $\beta = -9$

$a = -4 - 9 = -13$

At this point, we have established that the possible values of a are: -37, -20, -15 and -9.

The required sum is:

$Sum = -37 -20 -15 - 9$

$Sum = -81$

7 0

2 years ago

A quality control engineer at a potato chip company tests the bag filling machine by weighing bags of potato chips. Not every ba

Ahat [919]

Answer:

$z=\frac{0.21 -0.15}{\sqrt{\frac{0.15(1-0.15)}{100}}}=1.68$

Step-by-step explanation:

Information provided

n=100 represent the random sample taken

X=21 represent the number of bags overfilled

$\hat p=\frac{21}{100}=0.21$ estimated proportion of overfilled bags

$p_o=0.15$ is the value that we want to test

z would represent the statistic

Hypothesis

We need to conduct a hypothesis in order to test if the true proportion of overfilled bags is higher than 0.15.:

Null hypothesis: $p =0.7$

Alternative hypothesis: $p > 0.15$

The statistic for this case is:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

And replacing the info given we got:

$z=\frac{0.21 -0.15}{\sqrt{\frac{0.15(1-0.15)}{100}}}=1.68$

5 0

2 years ago