All categories

2 years ago

Ca someone plzzz help me with this problem?

Mathematics

1 answer:

Tamiku [17]2 years ago

5 0

So the answer to this would have to be worked out.

y = -7x + 1

You have two points, use those for the y and x.

1 = -7(0) + 1

1 = 1

The answer is yes.

You might be interested in

5 1/4 divided by 1/3 =

shepuryov [24]

I think it's 1 3/4. I had one similar earlier.

5 0

2 years ago

Read 2 more answers

HELP PLS IM FAILING PYTHAGOREAN THEOREM

Bond [772]

Answer: 4.6 km

Step-by-step explanation:

a^1 + b^2 = c^2

2^2 + b^2 = 5^2

4 + b^2 = 25

b = sqrt 25 - 4

b = sqrt 21 = 4.6 km

4 0

2 years ago

Read 2 more answers

Use a model to divide. 1/4 ÷ 5 1/20 4/5 1 1/4 20

Lera25 [3.4K]

Answer:

$\frac{1}{4} \div 5 \\ = \frac{1}{20}$

Pls....can mark brainliest

3 0

2 years ago

Read 2 more answers

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

2 years ago

Alicia is writing the program for a video game.

ICE Princess25 [194]

Step-by-step explanation:

A.-6-3 = -9

0+4 = 4

(-9,4)

B.3-3 = 0

-4+4 = 0

(0,0)

7 0

2 years ago