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Marta_Voda [28]
3 years ago
15

I need help With this I will mark you if your awnser is right

Mathematics
1 answer:
ElenaW [278]3 years ago
8 0

Answer:

\boxed {\sf Equation: 60 \textdegree +4x= 180 \textdegree} \\

\boxed {\sf x=30}

Step-by-step explanation:

The interior angles of a triangle must add to 180 degrees.

In this triangle, the angles are A, B , and C. So,

  • ∠A + ∠B + ∠C= 180°

∠A is 60 degrees, ∠B is 3x, and ∠C is x. We can substitute the values in.

  • 60°+3x+x=180°

This is the equation we can use to find the value of x.

We also have like terms: 3x and x that can be combined.

  • 60° + (3x+x)= 180°
  • <u>60° + 4x=180° </u>

<u />

We can use the equation to solve for x. We have to isolate the variable.

60 is being added to 4x. The inverse of addition is subtraction. Subtract 60 from both sides of the equation.

  • 60°-60°+4x=180°-60°
  • 4x= 180°-60°
  • 4x=120°

x is being multiplied by 4. The inverse of multiplication is division. Divide both sides by 4.

  • 4x/4= 120°/4
  • x= 120°/4
  • <u>x=30 </u>

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allsm [11]

Answer:

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3 0
3 years ago
Please help I will reward Brainly
grigory [225]
V=(4/3)(pi)(r^3)
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7 0
3 years ago
Sean took 4 hours to travel from Town A to Town B at an
labwork [276]

Answer:

Tina's average speed  for the whole journey = 56 kmph

Step-by-step explanation:

Time taken by Sean to travel from Town A to Town B = 4 hours.

Average speed of Sean = 70 km/h

We have equation of motion, s = ut + 0.5 at²

Time, t = 4 hours.

Initial speed, u = 70 km/hr

Acceleration, a = 0 m/s²

Substituting

      s = ut + 0.5 at²

      s = 70 x 4 + 0.5 x 0 x  4²

       s = 280 km

Distance from Town A to Town B = 280 km

Tina took 1 hour more than Sean.

Time taken by Tina to travel from Town A to Town B = 4 +1 = 5 hours

We have equation of motion, s = ut + 0.5 at²

Time, t = 5 hours.

Displacement, s = 280 km

Acceleration, a = 0 m/s²

Substituting

      s = ut + 0.5 at²

      280 = u x 5 + 0.5 x 0 x  5²

       u = 56 km/hr

Tina's average speed  for the whole journey = 56 kmph

6 0
3 years ago
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0
3 years ago
Plz help me out!!!!!!!!!
ad-work [718]

between 3 and 4 would be 3.5cm

8 0
3 years ago
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