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mafiozo [28]
2 years ago
11

Find the value xthis is a test I need it immediately asap​

Mathematics
2 answers:
-BARSIC- [3]2 years ago
7 0

Answer:

It should be 98 degrees

Step-by-step explanation:

All angles in a triangle add to be 180

So 180 -57 = 123

123 - 41 = 82

This gives you the missing interior angle

Now, the angle of a straight line is 180

so, 180 - 82 = 98

mamaluj [8]2 years ago
5 0

Answer:

98

Step-by-step explanation:

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Someone plz help me
topjm [15]

x = 375 and y = 6300. Option C.

Step-by-step explanation:

From the given data we need to find the value of x and y.

First, we will find out the ratio of area and the force.

\frac{Area}{Force}  = \frac{125}{1875} =\frac{150}{2250} =\frac{175}{2625} = \frac{1}{15}

So,

\frac{x}{5625} = \frac{1}{15}

or, x = \frac{5625}{15} = 375

Again,

\frac{420}{y} = \frac{1}{15}

or, y = 420×15 = 6300

Hence,

x = 375 and y = 6300.

7 0
3 years ago
A filter filled with liquid is in the shape of a vertex-down cone with a height of 9 inches and a diameter of 6 inches at its op
Alina [70]

Answer: Level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

Step-by-step explanation:

Since we have given that

Height = 9 inches

Diameter = 6 inches

Radius = 3 inches

So, \dfrac{r}{h}=\dfrac{3}{9}=\dfrac{1}{3}\\\\r=\dfrac{1}{3}h

Volume of cone is given by

V=\dfrac{1}{3}\pi r^2h\\\\V=\dfrac{1}{3}\pi \dfrac{1}{9}h^2\times h\\\\V=\dfrac{1}{27}\pi h^3

By differentiating with respect to time t, we get that

\dfrac{dv}{dt}=\dfrac{1}{27}\pi \times 3\times h^2\dfrac{dh}{dt}=\dfrac{1}{9}\pi h^2\dfrac{dh}{dt}

Now,  the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second, ie \dfrac{dv}{dt}=-4\ in^3

and h = 2 inches deep.

-4=\dfrac{1}{9}\times \pi\times (2)^2\dfrac{dh}{dt}\\\\-9\pi =\dfrac{dh}{dt}\\\\-28.28=\dfrac{dh}{dt}

Hence, level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

7 0
2 years ago
This problem has been solved!See the answerA municipal bond service has three rating categories (A, B, and C). Suppose that in t
mariarad [96]

Answer:

a. \frac{35}{51}

b. \frac{51}{100}

c. \frac{1}{5}

Step-by-step explanation:

Suppose cities represented by C', suburbs represented by S and rural represented by R,

Let x be the total number of bonds issued throughout the US,

According to the question,

n(A) = 70% of x = 0.7x,

n(B) = 10% of x = 0.1x,

n(C) = 20% of x = 0.2x,

n(A∩C') = 50% of n(A) = 0.5 × 0.7x = 0.35x,

n(A∩S) = 20% of n(A) = 0.2 × 0.7x = 0.14x,

n(A∩R) = 30% of n(A) = 0.3 × 0.7x = 0.21x,

n(B∩C') = 40% of n(B) = 0.4 × 0.1x = 0.04x,

n(B∩S) = 30% of n(B) = 0.3 × 0.1x = 0.03x,

n(B∩R) = 30% of n(B) = 0.3 × 0.1x = 0.03x,

n(C∩C') = 60% of n(C) = 0.6 × 0.2x = 0.12x,

n(C∩S) = 15% of n(C) = 0.15 × 0.2x = 0.03x,

n(C∩R) = 25% of n(C) = 0.25 × 0.2x = 0.05x,

n(C') = n(A∩C')  + n(B∩C')  + n(C∩C')  = 0.35x + 0.04x + 0.12x = 0.51x

n(S) = n(A∩S) + n(B∩S) + n(C∩S) = 0.14x + 0.03x + 0.03x = 0.20x

a. The probability that it will receive an A rating, if a new municipal bond is to be issued by a city,

P(\frac{A}{C'})=\frac{P(A\cap C')}{P(C')}=\frac{0.35x/x}{0.51x/x}=\frac{0.35}{0.51}=\frac{35}{51}

b. The proportion of municipal bonds are issued by cities = \frac{n(C')}{x}

=\frac{0.51x}{x}

=\frac{51}{100}

c. The proportion of municipal bonds are issued by suburbs = \frac{n(S)}{x}

=\frac{0.20x}{x}

=\frac{20}{100}

=\frac{1}{5}

3 0
2 years ago
last year the humane society took in 21,340 animals. If the humane society was open all 52 weeks that year. how many animals wer
dedylja [7]
The average intake rate is (21,340 animals)/(52 weeks) ≈ 410.4 animals/week
8 0
3 years ago
Which statement is not true about the pattern shown 2/3 , 4/6, 8/12, 16/24 , ...
natta225 [31]
They are multiplying by two .
8 0
2 years ago
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