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3 years ago

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Describe the location of the point having the following coordinates. positive abscissa, positive ordinate Quadrant I Quadrant II

Quadrant III Quadrant IV NEXT QUESTION ASK FOR HELP

1 answer:

asambeis [7]3 years ago

8 0

The abscissa is the x coordinate whereas the ordinate is the y coordinate. Both x and y are positive in the first quadrant only.

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A builder is building 5 new homes. Each home has a foyer measuring 9 feet by 18 feet. Laminate flooring for each foyer costs $26

telo118 [61]

Answer:

Step-by-step explanation:

just multiply 26.91 to what ever 9 times 18 is and then multiply that annswer to 5

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3 years ago

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Write the equation of the line that is perpendicular to y=2x-7 and passes through the point (6, 5)

Pani-rosa [81]

Answer:

B

Step-by-step explanation:

we are given a equation of a line

we want to figure out the equation of the perpendicular line passes through the <u>(</u><u>6</u><u>,</u><u>5</u><u>)</u><u> </u>points

in order to do so

recall that,

$\displaystyle m_{ \text{perpendicular}} = - \frac{1}{m}$

we got from our given equation that m=2

because equation of a line is y=mx+b

thus

$\displaystyle m_{ \text{perpendicular}} = - \frac{1}{2}$

remember that, when we want to figure out perpendicular line or parallel line we should the formula given by

$\displaystyle y - y_{1} = m(x - x_{1})$

since we got our perpendicular m is -½, $x_1=6$ and $y_1=5$ , substitute

$\displaystyle y - 5 = - \frac{1}{2} (x - 6)$

to get the perpendicular equation you should simplify the above equation to y=mx+b form

distribute -½:

$\displaystyle y - 5 = - \frac{1}{2} x + 3$

add 5 to both sides:

$\displaystyle y = - \frac{1}{2} x + 8$

hence,

our answer choice is B

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3 years ago

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Answer my four questions below and you will get 20 points.

Mice21 [21]

1) 1 + 1 = 2

2) 1 - 1 = 0

3) 1 * 1 = 1

4) 1 / 1 = 1

Hope this helps!

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Write an equation of a parabola with vertex at the origin and the given focus<br> (6,0)

Masteriza [31]

Answer:

Step-by-step explanation:

eq. of a parabola with vertex at origin and focus (a,0) is

y²=4ax

a=6

so y²=4×6×x

or y²=24 x

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3 years ago

(1+tan^2 A)cot A/cosec^2 A= tanA

Rufina [12.5K]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

1 + tan²A = sec²A , secA = $\frac{1}{cosA}$ , cosecA = $\frac{1}{sinA}$ , cotA = $\frac{cosA}{sinA}$ , tanA = $\frac{sinA}{cosA}$

Consider the left side

$\frac{(1+tan^2A)cotA}{cosec^2A}$

= $\frac{sec^2AcotA}{cosec^2A}$

= $\frac{\frac{1}{cos^2A}(\frac{cosA}{sinA}) }{cosec^2A}$

= $\frac{\frac{1}{cosAsinA} }{\frac{1}{sin^2A} }$

= $\frac{1}{cosAsinA}$ × sin²A

= $\frac{sinA}{cosA}$

= tanA = right side , thus proven

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3 years ago