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2 years ago

Find the x and y intercepts c) x + 3y - 12=0

1 answer:

7 0

To get x intercept：
let y=0，x➖12=0，x=12
so x intercept point （12，0）
To get y intercept：
let x=0，3y➖12=0，y=4
so y intercept point （0，4）

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Jane is training for a triathlon. After swimming a few laps, she leaves the health club and bikes 16 miles south. She then runs

lianna [129]

Answer:

The equation for the distance Jane's trainer bikes is $x=\sqrt{16^2+12^2}$ .

Step-by-step explanation:

We have attached diagram for your reference.

Given:

Distance traveled on bike towards south = 16 miles

Distance she ran towards west = 12 miles

We need to find distance Jane's trainer bikes.

Solution:

Let the distance Jane's trainer bike be 'x'.

Now we will assume it to be right angled triangle.

So by Pythagoras theorem which states that;

"Square of the third side is equal to the sum of square of the other two sides."

framing in equation form we get;

$x^2=16^2+12^2\\\\x=\sqrt{16^2+12^2}$

Hence the equation for the distance Jane's trainer bikes is $x=\sqrt{16^2+12^2}$ .

On solving we get;

$x=\sqrt{16^2+12^2}\\\\x= \sqrt{256+144}\\ \\x=\sqrt{400} \\\\x=20\ miles$

Hence Jane's trainer bikes a distance of 20 miles.

7 0

3 years ago

Can the values be added or subtracted? If one value needs to be converted to another, describe the conversion.

Nastasia [14]

C. the values can be added after making a conversion of units because they both measure length.

8 0

3 years ago

Fine length of BC on the following photo.

MrMuchimi

Answer:

$BC=4\sqrt{5}\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

$AC^2=AD^2+DC^2$

substitute the given values

$AC^2=16^2+8^2$

$AC^2=320$

$AC=\sqrt{320}\ units$

simplify

$AC=8\sqrt{5}\ units$

step 2

In the right triangle ACD

Find the cosine of angle CAD

$cos(\angle CAD)=\frac{AD}{AC}$

substitute the given values

$cos(\angle CAD)=\frac{16}{8\sqrt{5}}$

$cos(\angle CAD)=\frac{2}{\sqrt{5}}$ ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

$cos(\angle BAC)=\frac{AC}{AB}$

substitute the given values

$cos(\angle BAC)=\frac{8\sqrt{5}}{16+x}$ ----> equation B

step 4

Find the value of x

In this problem

$\angle CAD=\angle BAC$ ----> is the same angle

equate equation A and equation B

$\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}$

solve for x

Multiply in cross

$(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units$

$DB=4\ units$

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

$BC^2=DC^2+DB^2$

substitute the given values

$BC^2=8^2+4^2$

$BC^2=80$

$BC=\sqrt{80}\ units$

simplify

$BC=4\sqrt{5}\ units$

7 0

2 years ago

In this question, all lengths are in centimetres.

JulsSmile [24]

Answer:

Step-by-step explanation:

AB/AC =1/2

(3x-4)/(2x+12)=1/2

2(3x-4) = 1(2x+12)

6x-8=2x+12

4x=20

X=5

AC/BC =3x-4/7x-2=22/33=2/3

AB/AC=3x-4/7x-2=22/33=2/3

AB/AC=3x-4/2x+12

6x-8=2x+12

You would get 4/5 still trying to work out the last mark

3 0

3 years ago

A basket holds at most 14 pounds of apples and oranges

Helen [10]

A.

B and C are wrong because x is more than or equal to 3.

D is wrong because y is the weight of oranges and cannot be negative.

7 0

3 years ago

Read 2 more answers