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riadik2000 [5.3K]

2 years ago

9

You have a photo that measures 1 in. wide by 3 in. tall. If you want to enlarge the photo so that the height is 18 in., what wil

l the width of the photo be?

1 answer:

kiruha [24]2 years ago

4 0

6in... 3 times 6 is 18, so 1 times 6 is 6

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Are the following lines parallel? <br><br> y = -3x + 1 <br><br> y = 4 - 3x

PIT_PIT [208]

Answer:

Simple! The answer is yes.

Step-by-step explanation:

Simply put the equations into Desmos graphing calculator and it will show you that the equations are parallel. :))

4 0

2 years ago

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

5 0

3 years ago

What is the answer ?

adell [148]

The most reasonable would be grams.

5 0

2 years ago

Read 2 more answers

A teenager can wash 2

leva [86]

2 * M = X
M = to minutes
x = to how many tables washes

3 0

3 years ago

Find the HCF and LCM of the following pairs of numbers:

Marina86 [1]

Answer:

a)HCF between 774&252 is=18

LCM between 774&252=10836

b)HCF between 204&544 is=68

LCM between 204&544 is=1632

c)HCF between 770&385 is=385

LCM between 770&385 is =770

d)HCF between 477&552=3

LCM between 447&552 is =82248

Step-by-step explanation:

a)because LCM=2×2×3×3×7×43

b)because LCM=2×2×2×2×2×3×17

c)because LCM=2×5×7×11

d)because LCM=2×2×2×3×23×149

4 0

3 years ago