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Nuetrik [128]
3 years ago
12

Find the missing length marked with a question mark. UVW-USR

Mathematics
1 answer:
mariarad [96]3 years ago
3 0

Given:

\Delta UVW\sim \Delta USR

To find:

The missing length marked with a question mark.

Solution:

Let x be the missing value.

We have,

\Delta UVW\sim \Delta USR

Corresponding sides of similar triangles are proportional, so

\dfrac{UV}{US}=\dfrac{VW}{SR}=\dfrac{UW}{UR}

Using this, we get

\dfrac{UV}{US}=\dfrac{UW}{UR}

On substituting the values, we get

\dfrac{70}{x}=\dfrac{55}{44}

\dfrac{70}{x}=\dfrac{5}{4}

On cross multiplication, we get

70\times 4=5\times x

280=5x

\dfrac{280}{5}=x

56=x

Therefore, the missing length is 56 units.

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Lisa drew three circles to form a figure. The areas of the circles were in the
Tamiku [17]

Answer:

Fraction of the figure shaded = \frac{13}{16}

Step-by-step explanation:

Ratio of the areas of the given circles are 1 : 4 : 16

Then the radii of the circles will be in the ratio = \sqrt{1}:\sqrt{4}:\sqrt{16}

                                                                             = 1 : 2 : 4

If the radius of the smallest circle = x units

Then the radius of the middle circle = 2x units

and the radius of the largest circle = 4x units

Area of the smallest circle = πx²

Area of the middle circle = π(2x)² = 4πx²

Area of the largest circle = π(4x)²= 16πx²

Area of the region which is not shaded in the middle circle = πx²(4 - 1)

                                                                                                  = 3πx²

Therefore, area of the shaded region = Area of the largest circle - Area of the region which is not shaded

= 16πx² - 3πx²

= 13πx²

Fraction of the figure which is not shaded = \frac{\text{Area of the shaded region}}{\text{Area of the largest circle}}

= \frac{13\pi x^{2} }{16\pi x^{2} }

= \frac{13}{16}

6 0
3 years ago
Please solve this 4 the grade algebra problem? A group of students calculated their average score at a spelling bee. They realiz
In-s [12.5K]
Let x = original average score
let y = be the number of students

We need to write two equation using the information given in the scenario in order to work them simultaneously and obtain the results.

let \frac{yx  +   9}{y}  =  81 .... (1)
     so x is the original average, so we multiply that average by the amount of     students [y × x] in order to obtain their cumulative score then you add the       nine to that score (because a student got 9 more points) [yx + 9]. Then you    divide that sum by the amount of student in  order to get the new average      which the question says would be 81

\frac{yx  -  3}{y}  =  78 ...... (2)
   so x is the original average, so we multiply that average by the amount of     students [y × x] in order to obtain their cumulative score then you subtract       three from that score since one student got three less points [yx - 3].  Then     you divide by the number of students (y) and you should 78 like the                 question says.

\frac{yx + 9}{y} = 81 .... (1)
\frac{yx - 3}{y} = 78 ...... (2)

Now simplify each equation by separating the LHS
\frac{yx}{y}  +   \frac{9}{y}  = 81..... (1a)
\frac{yx}{y} - \frac{3}{y} = 78...... (2a)

By subtracting eq (2a) from eq (1a) in order to eliminate x
\frac{yx}{y} - \frac{yx}{y} + \frac{9}{y} - (- \frac{3}{y}) = 81 - 78

\frac{12}{y} = 3

\frac{12}{3} = y

⇒ y  =  4

Since y = the number of students
then  the number of students = 4




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The length of a rectangular swimming pool is twice its width minus 3 meters. If x represents the width of the swimming pool in m
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