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aleksandrvk [35]
2 years ago
11

please help I'm not good at math. What is the equation of the line that has a slope of -4 and passes through the point (2, 3)?

Mathematics
2 answers:
harkovskaia [24]2 years ago
5 0

Answer

y=-4x-1

Step-by-step explanation:

Katen [24]2 years ago
5 0
Y=-4x1 hope this help
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What is the volume of this solid? Recall the formula V = Bh.
Viktor [21]

Answer:

280

Step-by-step explanation:

V=Bh

V=1/2(7x10)8      *it is a triangluar pyrmid so the base is a triangle*

V=1/2x70x8

V=35x8

V=280

Answer:280

5 0
3 years ago
Read 2 more answers
A line had a slope of 2/3 and a y intercept at (0,14).What is the x intercept of the line?​
pychu [463]

Answer:

y=2/3x+14

Step-by-step explanation:

The standard form of an equation in slope-intercept form is y=mx+b where m=slope and b=y-intercept.

Given a y-intercept of 14 and a slope of 2/3, we can plug into the variables and get the equation y=2/3x+14

The x-intercept would be when y=0, so plugging in y=0 to the equation gets us:

0=2/3x+14

-14=2/3x

21=x

So the x-intercept is 21

6 0
2 years ago
Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter
Levart [38]

Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

.............x/sqrt(64+x^2)

.............sqrt(64-x^2)/8

..............x/sqrt(64-x^2)

..............8/sqrt(64+x^2)

A. ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and  \sqrt{8^{2} - x^{2} }or  \sqrt{64 - x^{2} }

thus tan(arcsin(x/8))  = \frac{x}{\sqrt{64 - x^{2} }}  =

Therefore  ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B

Here we have cos = adjacent/hypotenuse where adjacent side is \sqrt{64 - x^{2} } and hypothenuse = 8 we have \sqrt{64 - x^{2} }/8

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

4 0
2 years ago
Help please answer these question
yawa3891 [41]

Answer:

#3. d = -10.25

Step-by-step explanation:

#3. d = -6.5 + -3.75

8 0
3 years ago
XPF is congruent to which correctly named triangle
guapka [62]

Answer:

the answer is triangle GBL

7 0
3 years ago
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