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2 years ago

13

After leaving her house, Shelley drove 14.47 kilometers to the gas station and then another 5.8 kilometers to work. How many tot

al kilometers did Shelley drive to get to work after she left her house? Show your work on paper. Then, enter your answer in decimal form in the box

1 answer:

shutvik [7]2 years ago

5 0

Answer:

20.27 km

Step-by-step explanation:

14.47

+ <u> 5.8</u>

<u> </u> 20.27

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2.1 ÷ 1.488 round to nearest tenth

Lorico [155]

I believe it should be 1.4 since the answer you get is 1.4112903226... you would round but on the tenths place 1.4

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2 years ago

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3. How many 1/8's are in 1 1/4?

Murrr4er [49]

Answer:

10

Step-by-step explanation:

There are two 1/8 in one 1/4. There are eight 1/8 in 1. 8/8 + 1/8 + 1/8 = ten 1/8.

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2 years ago

Oints Q and R are midpoints of the sides of triangle ABC.

MariettaO [177]

Answer:

AQ = 20 units

Step-by-step explanation:

Comparing triangle AQR to ABC,

$\frac{AQ}{AB}$ = $\frac{QR}{BC}$

$\frac{4p}{8p}$ = $\frac{2p + 3}{6p-4}$

cross multiply and make p the subject of formula, we have:

8p (2p+3) = 4p(6p-4)

16 $p^{2}$ + 24p = 24 $p^{2}$ - 16p

24p + 16p = 24 $p^{2}$ - 16 $p^{2}$

40p = 8 $p^{2}$

divide through by 8p,

p = 5

Therefore, AQ = 4p

= 4 × 5

= 20

AQ = 20 units

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2 years ago

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

Solve the equation, and check the solutions (√x+3)+6=3.

larisa86 [58]

Answer:

The equation has no solution.

Step-by-step explanation:

Hi there!

First, let´s write the equation:

(√x + 3) + 6 = 3

To solve the equation, let´s subtract 6 to both sides of the equation:

√x + 3 = -3

subtract 3 to both sides of the equation:

√x = -6 (the square root of any number can´t be negative, but let´s solve it anyway and then check the solution).

Square both sides of the equation:

x = 36

Now, let´s check the solution:

(√x + 3) + 6 = 3

x = 36

(√36 + 3) + 6 = 3

(6 + 3) + 6 = 3

9 + 6 = 3

15 ≠ 3

Then, the equation has no solution.

Have a nice day!

6 0

2 years ago