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3 years ago

Use the equation of a line with a slope of 4 and a y-intercept of -3.

Mathematics

1 answer:

GarryVolchara [31]3 years ago

4 0

Hello
the answer is c

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Anyone know what this means

qaws [65]

Answer:

Positive exponent

Step-by-step explanation:

Like x root 1 , 2 , 3 etc

If it said negative exponent then x root -1, -2 etc etc.

6 0

2 years ago

Read 2 more answers

Where does the circumference of each kind of triangle lie? Question 1 options: Inside the triangle. Outside the triangle. On the

stiks02 [169]

Answer:

Inside the triangle - Acute Triangle

Outside the triangle - Obtuse Triangle

On the hypotenuse - Right Triangle

Step-by-step explanation:

The circumcenter is the point where the perpendicular bisectors of a triangle intersect.

In the special case of a right triangle, the circumcenter lies exactly at the midpoint of the hypotenuse.

The circumcenter of an acute triangle lies inside the triangle.

The circumcenter of an obtuse triangle lies outside the triangle.

4 0

2 years ago

Help with substitution show work please!!!!!!!

yaroslaw [1]

Answer: x=4, y=-2

Step-by-step explanation: 4x+y=14, x-2y=8

4x+y-y=14-y

4x/4=14/4-y/4

x=14-y/4-(2y)(4)/4

4(14-9y)/4=(8)(4)

14-9y-14=32-14

-9y/-9=18/-9

y=-2

(14)-(-2)/4

14+2/4

16/4

x=4

x=4, y=-2

6 0

2 years ago

Read 2 more answers

Given: △ABC, AB=5 square root 2 m∠A=45°, m∠C=30° Find: BC and AC

hichkok12 [17]

Answer:

Part a) $BC=10\ units$

Part b) $AC=13.66\ units$

Step-by-step explanation:

step 1

Find the length side BC

Applying the law of sines

we know that

$\frac{AB}{sin(C)}=\frac{BC}{sin(A)}$

we have

$AB=5\sqrt{2}\ units$

$A=45^o$

$C=30^o$

substitute

$\frac{5\sqrt{2}}{sin(30^o)}=\frac{BC}{sin(45^o)}$

solve for BC

$BC=\frac{5\sqrt{2}}{sin(30^o)}(sin(45^o))$

$BC=10\ units$

step 2

Find the measure of angle B

we know that

The sum of the interior angles in a triangle must be equal to 180 degrees

$m\angle A+m\angle B+m\angle C=180^o$

substitute the given values

$45^o+m\angle B+30^o=180^o$

$75^o+m\angle B=180^o$

$m\angle B=180^o-75^o$

$m\angle B=105^o$

step 3

Find the length side AC

Applying the law of sines

we know that

$\frac{AB}{sin(C)}=\frac{AC}{sin(B)}$

we have

$AB=5\sqrt{2}\ units$

$A=45^o$

$B=105^o$

substitute

$\frac{5\sqrt{2}}{sin(30^o)}=\frac{AC}{sin(105^o)}$

solve for AC

$AC=\frac{5\sqrt{2}}{sin(30^o)}(sin(105^o))$

$AC=13.66\ units$

4 0

2 years ago

Odelia earns $16,750 and has $2,870 withheld for taxes and $800 with-held for personal deductions. If she gets paid biweekly, wh

slega [8]

$16,750 -$2,870 -$800 = $13,080

Biweekly means she is paid 26 times a year.
$13,080 ÷ 26 = 503.0769231

b. $503.08

6 0

2 years ago