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grandymaker [24]

3 years ago

10

What did I do wrong? It is a linear equation.

1 answer:

iren2701 [21]3 years ago

6 0

You only did the last one wrong, (3/4) times 2 is 3/2. 3/2 - 1/2 is 2/2 or 1.

So it'd be (2,1)

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If a man runs for an hour and a half at 8 miles an hour due south; and then he continues south on a bus traveling at 55 miles an

Usimov [2.4K]

He ran 8 miles per hour.

Multiply his speed by the time he ran to find total distance:

8 mph x 1.5 hours = 12 miles

He ran for 12 miles.

Now Find the distance the bus went:

For time, divide the minutes the bus drove by 60 minutes per hour.

55 miles per hour x 24/60 = 22 miles

The bus went 22 miles.

Total distance = 12 + 22 = 34 miles.

4 0

3 years ago

How do i do x^2 - 10x = -25

Tatiana [17]

Set the single term factor on the left side of the equation equal to 0 0 . x=0 x = 0.
Set each of the factors of the left side of the equation equal to 0 0 . ...
Since −10 - 10 does not contain the variable to solve for, move it to the right side of the equation by adding 10 10 to both sides. ...
Solve the equations for x x .

3 0

3 years ago

Can anyone do 4,5 and 6 for me plz

cricket20 [7]

For question 4, $sinA =\frac{a}{c}, cosA= \frac{b}{c}, tan B = \frac{b}{a}, sin J = \frac{j}{l}, cosK = \frac{j}{l}, tanK = \frac{k}{j}.$

Question 5. Option a and question 6. Option j

Step-by-step explanation:

Step 1:

The three basic formula needed to solve these questions are:

$sin\theta = \frac{oppositeside}{hypotenuse} , cos\theta = \frac{adjacentside}{hypotenuse}, tan\theta= \frac{opposite side}{adjacent side}.$

Step 2:

Using the above formula, we solve the following values

$sinA = \frac{oppositeside}{hypotenuse}$ $=\frac{a}{c}.$

$cosA = \frac{adjacentside}{hypotenuse}$ $= \frac{b}{c}.$

$tanB= \frac{opposite side}{adjacent side}$ $= \frac{b}{a}.$

$sinJ = \frac{oppositeside}{hypotenuse}$ $= \frac{j}{l}.$

$cosK = \frac{adjacentside}{hypotenuse}$ $= \frac{j}{l}.$

$tanK= \frac{opposite side}{adjacent side}$ $= \frac{k}{j}.$

Step 3:

For question 5, The triangle's angle = 23°, opposite side = BC inches and hypotenuse = 4 inches.

$sin\theta= \frac{opposite side}{hypotenuse}. sin 23^{\circ}= \frac{BC}{4}, sin23^{\circ} = 0.3907,BC = (0.3907)(4) = 1.5628.$

SO BC is 1.5628 inches, rounding this off to the nearest tenth, we get BC = 1.6 inches which is option a.

Step 4:

For question 6, The triangle's angle = 50°, opposite side = QR m the adjacent side = 8.1 m.

$tan\theta= \frac{opposite side}{adjacentside}. tan 50^{\circ}=\frac{QR}{8.1}, tan50^{\circ} = 1.1917,QR = (1.1917)(8.1) = 9.65277$

SO QR is 9.65277 meters, rounding this off to the nearest tenth, we get QR = 9.7 inches which is option j.

7 0

3 years ago

How can you model 8÷2/7

fredd [130]

8÷2/7
8/1÷2/7
8/1*7/2
56/2
28

Hope this helps :)

3 0

4 years ago

Brian found 12-19 by breaking apart 19 into 12+7 write equations to show how Brian could have found the difference. ?

melomori [17]

Answer:

Answer is 12-19 = -7

Step-by-step explanation:

12- 19 you can break 19 into 12 +7

then

it is easy to find 12-12 = 0

Now subtracts 0 - 7 =- 7

then

break 7 into 3 + 4

then

0 - 3 = -3

-3 - 4 = -7

So ,

12 - 19 = -7

5 0

3 years ago