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3 years ago

13

Please include Work!!!

1 answer:

Natalka [10]3 years ago

8 0

5.3/8=.6625 yrds^2 should be right

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What is the equation of the line? Write in standard form. (4,5).

dezoksy [38]

Answer:

$\large\boxed{5x+4y=20}$

Step-by-step explanation:

The equation of a line in slope-intercept form:

$y=mx+b$

m - slope

b - y-intercept → (0, b)

We have the x-intercept = 4 → (4, 0), and y-intercept = 5 → b = 5.

Therefore we have the equation:

$y=mx+5$

Put the coordinates of x-intercept to the equation:

$0=4m+5$ <em>subtract 4m from both sides</em>

$-4m=5$ <em>divide both sides by (-4)</em>

$m=-\dfrac{5}{4}$

Finally we have the equation of a line if slope-intercept form:

$y=-\dfrac{5}{4}x+5$

The standard form:

$Ax+By=C$

Convert:

$y=-\dfrac{5}{4}x+5$ <em>multiply both sides by 4</em>

$4y=-5x+20$ <em>add 5x to both sides</em>

$5x+4y=20$

4 0

3 years ago

an exterminator must treat the perimeter of a house for insects if the house is 90 feet long and 30 feet wide what is the total

Andru [333]

240 feet. hope this helped

6 0

4 years ago

the triangles below are congruent and their corresponding parts are marked. name all the corresponding congruent angles and side

GuDViN [60]

Answer:

See explanation

Step-by-step explanation:

The triangles below are congruent and their corresponding parts are marked.

a) Angles marked with one arc are congruent, so $\angle B\cong \angle Z.$

Angles marked with two arcs are congruent, so $\angle A\cong \angle Y.$

Angles marked with three arcs are congruent, so $\angle C\cong \angle X.$

b) Sides marked with one segment are congruent, so $\overline {AC}\cong \overline {YX}.$

Sides marked with two segments are congruent, so $\overline {BC}\cong \overline {ZX}.$

Sides marked with three segments are congruent, so $\overline {AB}\cong \overline {YZ}.$

c) Name teo congruent triangle following the congruence of angles:

$\triangle BCA\cong \triangle ZXY$

6 0

4 years ago

Find the area of quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21)

sasho [114]

Answer:

The area of quadrilateral ABCD is 139 unit^2.

Step-by-step explanation:

Given:

Quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21).

Now, to find the area.

The coordinates of the quadrilateral are A(1,1), B(7,-3), C(12,2), D(7,21).

So, to find the area we need to bisect the quadrilateral ABCD and get the triangles ABC and ADC and then calculate their areas:

In Δ ABC:

$A(x_1,y_1)=(1,1)\:,\:B(x_2,y_2)=(7,-3)\:and\:C(x_3,y_3)=(12,2)$

Now, to get the area of triangle ABC:

$Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-3-2)+(7)(2-1)+12(1--3)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-5)+(7)(1)+12(4)\right|$

On solving we get:

$Area\,of\,triangle\,=25.$

In Δ ADC:

$A(x_1,y_1)=(1,1)\:,\:D(x_2,y_2)=(7,21)\:and\:C(x_3,y_3)=(12,2)$

Now, to get the area of triangle ADC:

$Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(21-2)+(7)(2-1)+12(1-21)\right|$

On solving it by same process as above we get:

$Area\,of\,triangle\,=114$

Now, to get the area of the quadrilateral we add the areas of the triangles ABC and ADC:

$25+114\\=25+114\\=139\ unit^2$

Therefore, area of quadrilateral ABCD is 139 unit^2.

4 0

3 years ago

1. You flip a coin four times. What is the probability that all four of them are heads? and 2. You pick three cards from a deck

S_A_V [24]

1. you have a 50 50 chance

2. 4/52 chance

Reason

1. coins only have two sides

2. there are 4 kings in a deck of 52 cards

6 0

3 years ago