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Olenka [21]
3 years ago
6

How do you solve this equation 7(3)+(-9)×(-4)=

Mathematics
1 answer:
Softa [21]3 years ago
7 0

Hello!

To solve this equation, we will need to use PEMDAS (parentheses, then exponents, then multiplication/division, then addition/subtraction).

7(3) + (-9) x (-4) =

21 + (-9) x (-4) =

21 + 36 =

57

I hope this helps you! Have a lovely day!

- Mal

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What is the unit rate for 14 lb for $2.99
Alexeev081 [22]
How much for 1lb
2.99/14=0.213/1=$0.21

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3 years ago
What is the best estimate of the perimeter of the figure on the grid if each square has side lengths of 1 mm?
Wewaii [24]

Answer:

4mm

Step-by-step explanation:

you have to add both sides after substituting each of the 4 sides by 1mm

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6 0
1 year 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
Solving systems of equations using substitution p=q+3 4p+3q= -30
patriot [66]
4(q+3) + 3q = -30
4q +12 + 3q = -30
7q+12= -30
7q = -12-30
7q = -42
q = -6
p = -3

5 0
3 years ago
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