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IRINA_888 [86]
3 years ago
14

Okayy but fr someone helppp, plss ASAP. I need an answer now

Mathematics
1 answer:
hjlf3 years ago
5 0

Answer:

ertyuiygguyyguguyugggu

Step-by-step explanation:

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Macy's final grades are a A, B, C, D, D, D, F, all are weighed the same, what would her final grade be?
Inga [223]
Her final grade would be 70%
6 0
3 years ago
A school typically sells 500 yearbooks each year for $50. The economics class does a project and discovers that they can sell 10
padilas [110]

Answer:

a is 100 b is 5

Step-by-step explanation:

a is 100 because if you substitute 100 for a and 1 for x (decrease of 5 dollars) you get 600 yearbooks sold which is a curate according to the economics class prediction

b is 5 because substituting 5 into b and 1 to x gives you 45 dollars per year book. this is accurate because a 5.dollar decrease in price makes the year books 45 dollars instead of 50

7 0
3 years ago
Enter the ordered pair for the vertices for Rx-axis(QRST).
mash [69]

Given:

The vertices of a polygon QRST are T(-2, 3), Q(1, 5), R(3, -1) and S(0, 0).

To find:

The vertices for R_{x-axis}(QRST).

Solution:

The rule R_{x-axis}(QRST) represents refection of polygon QRST across the x-axis.

If a figure is reflected across the x-axis, then

(x,y)\to (x,-y)

Using this rule, we get

T(-2,3)\to T'(-2,-3)

Q(1,5)\to Q'(1,-5)

R(3,-1)\to R'(3,1)

S(0,0)\to S'(0,0)

Therefore, the vertices for R_{x-axis}(QRST) are T'(-2, -3), Q'(1, -5), R'(3, 1) and S'(0, 0).

7 0
3 years ago
The difference between the observed value of the dependent variable and the value predicted using the estimated regression equat
Elenna [48]

Answer:

For this case we define the dependent variable as Y and the independent variable X. We assume that we have n observations and that means the following pairs:

(x_1, y_1) ,....,(x_n,y_n)

For this case we assume that we want to find a linear regression model given by:

\hat y = \hat m x +\hat b

Where:

\hat m represent the estimated slope for the model

\hat b represent the estimated intercept for the model

And for any estimation of the dependent variable \hat y_i , i=1,...,n is given by this model.

The difference between the observed value of the dependnet variable and the value predicted using the estimated regression equation is known as residual, and the residual is given by this formula:

e_i = y_i -\hat y_i , i=1,...,n

So the best option for this case is:

d. residual

Step-by-step explanation:

For this case we define the dependent variable as Y and the independent variable X. We assume that we have n observations and that means the following pairs:

(x_1, y_1) ,....,(x_n,y_n)

For this case we assume that we want to find a linear regression model given by:

\hat y = \hat m x +\hat b

Where:

\hat m represent the estimated slope for the model

\hat b represent the estimated intercept for the model

And for any estimation of the dependent variable \hat y_i , i=1,...,n is given by this model.

The difference between the observed value of the dependnet variable and the value predicted using the estimated regression equation is known as residual, and the residual is given by this formula:

e_i = y_i -\hat y_i , i=1,...,n

So the best option for this case is:

d. residual

7 0
3 years ago
Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A
mash [69]

Answer:

Step-by-step explanation:

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA

we take the LHS so here goes,

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\

since , sin^2A+cos^2A=1

the identity becomes,

(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\

now, we know,

sin^2A=1-cos^2A and cos^2A=1-sin^2A

the identity becomes,

(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\

(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})

sin A and cos A cancel out it becomes zero

\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\

Splitting the denominator the identity becomes

\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A}  \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\

Hence,

tanAsecA-cotAcosecA

3 0
3 years ago
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