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3 years ago

How does the distributive property relate to the factoring expression ?

1 answer:

3 0

Factoring – Using the Distributive Property. A factor is a number that can be divided into another number evenly. For example, the factors of 6 are 1, 2, 3, and 6. ... It means that a number outside the parentheses of an addition problem can be multiplied by each number inside the parentheses.

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A county fair charges $5 for tickets and $2 for each ride. What is the y-intercept in this scenario?

zhenek [66]

Answer:

(I think) ¯\_(ツ)_/¯

Step-by-step explanation:

cause 5 would be X and then that leaves 2 for Y

(5,2)

(X,Y)

6 0

3 years ago

Read 2 more answers

Which ordered pairs are in a proportional relationship with (0.2, 0.3)?

MAXImum [283]

Answer:

C and E

Step-by-step explanation:

We are given that

$x_1=0.2,y_1=0.3$

$\frac{y}{x}=\frac{0.3}{0.2}=\frac{3}{2}$

$k=\frac{y}{x}=\frac{3}{2}$

A. $x_2=1.2,y_2=2.3$

$\frac{y_2}{x_2}=\frac{2.3}{1.2}=\frac{23}{12}\neq=\frac{3}{2}$

Hence, it is not in proportional relationship with (0.2,0.3)

B.(2.7,4.3)

$x_3=2.7,y_3=4.3$

$\frac{y_3}{x_3}=\frac{4.3}{2.7}=\frac{43}{27}\neq\frac{3}{2}$

Hence, it is not in proportional relationship with (0.2,0.3).

C.(3.2,4.8)

$x_4=3.2,y_4=4.8$

$\frac{y_4}{x_4}=\frac{4.8}{3.2}=\frac{3}{2}$

Hence, the ordered pair (3.2,4.8) are in a proportional relationship with (0.2,0.3).

D.(3.5,5.3)

$\frac{y_5}{x_5}=\frac{5.3}{3.5}=\frac{53}{35}\neq \frac{3}{2}$

Hence, the ordered pair (3.5,5.3) are not in a proportional relationship with (0.2,0.3).

E.(5.2,7.8)

$\frac{y_6}{x_6}=\frac{7.8}{5.2}=\frac{3}{2}$

Hence, the ordered pair (5.2,7.8) are in a proportional relationship with (0.2,0.3).

4 0

3 years ago

Could you give me the steps to find the value of x in this figure? Worth 11 points!

vladimir2022 [97]

The angles x and 55.1 degrees are COMPLEMENTARY ANGLES. Please look up that term if necessary.

Subtract the angle 55.1 degrees from ( what? ) to obtain angle x.

6 0

4 years ago

Write the equation in standard form. Identity the center and radius.<br> x² + y2 + 8x-4y-7=0

lorasvet [3.4K]

Answer:

equation;

$(x + 4) {}^{2} + (y - 2) {}^{2} = 27$

Center (-4,2)

Radius is

$3 \sqrt{3}$

Step-by-step explanation:

Since the x^2 and y^2 have the same coeffiecent this will be a circle in a form of

$(x - h) {}^{2} + (y - k) {}^{2} = {r}^{2}$

Where (h,k) is center

r is the radius

So first we group like Terms together

${x}^{2} + 8x + {y}^{2} - 4y - 7 = 0$

Add 7 to both sides

${x}^{2} + 8x + {y}^{2} - 4y = 7$

$( {x}^{2} + 8x) +( {y}^{2} - 4y) = 7$

Since the orginal form of the equation of the circle has a perfect square we need to complete the square for each problem

$(\frac{8}{2} ) {}^{2} = 16$

and

$( - \frac{4}{2} ) {}^{2} = 4$

so we have

${x}^{2} + 8x + 16 + {y}^{2} - 4y + 4y = 7 + 16 + 4$

${x}^{2} + 8x + 16 + {y}^{2} - 4y + 4y = 27$

$(x + 4) {}^{2} + (y - 2) {}^{2} = 27$

To find our center, h is -4 and k is 2

so the center is (-4,2)

The radius is

$\sqrt{27} = 3 \sqrt{3}$

So the radius is 3 times sqr root of 3.

8 0

2 years ago

A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The

Oksanka [162]

Answer:

a) $[-0.134,0.034]$

b) We are uncertain

c) It will change significantly

Step-by-step explanation:

a) Since the variances are unknown, we use the t-test with 95% confidence interval, that is the significance level = 1-0.05 = 0.025.

Since we assume that the variances are equal, we use the pooled variance given as

$s_p^2 = \frac{ (n_1 -1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}$ ,

where $n_1 = 40, n_2 = 30, s_1 = 0.16, s_2 = 0.19$ .

The mean difference $\mu_1 - \mu_2 = 10.85 - 10.90 = -0.05$ .

The confidence interval is

$(\mu_1 - \mu_2) \pm t_{n_1+n_2-2,\alpha/2} \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}} = (-0.05) \pm t_{68,0.025} \sqrt{\frac{0.03}{40} + \frac{0.03}{30}}$

$= -0.05\pm 1.995 \times 0.042 = -0.05 \pm 0.084 = [-0.134,0.034]$

b) With 95% confidence, we can say that it is possible that the gaskets from shift 2 are, on average, wider than the gaskets from shift 1, because the mean difference extends to the negative interval or that the gaskets from shift 1 are wider, because the confidence interval extends to the positive interval.

c) Increasing the sample sizes results in a smaller margin of error, which gives us a narrower confidence interval, thus giving us a good idea of what the true mean difference is.

6 0

4 years ago

How does the distributive property relate to the factoring expression ?​

How does the distributive property relate to the factoring expression ?