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PtichkaEL [24]
4 years ago
13

A school contains 357 boys and 323 girls.

Mathematics
1 answer:
PolarNik [594]4 years ago
3 0

Answer:

19/40

Step-by-step explanation:

just divide the number of girls by the total number of students.

323/680 = 19/40 or 0.48

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Your cat weighs 10.4 pounds. How many kilograms does your cat weigh?
gavmur [86]

Answer:

The cat weighs 4.717 kg

One kilogram (kg) is 2.2 lbs

Divide 10.4 by 2.2

Hope this helps! :)

5 0
3 years ago
What is the approximate angle measure for angle W in the triangle below?
Artyom0805 [142]
In this case we know the three sides of the triangle, then this is a SSS triangle (Side Side Side). To solve this case, first we must use the Law of Cosines, applied to the opposite side to the angle we want to find.
We want to find angle W, and its opposite side is XV, then we apply the Law of Cosines to the side XV:
XV^2=XW^2+WV^2-2(XW)(WV)cos W
Replacing the known values:
116^2=96^2+89^2-2(96)(89)cos W
Solving for W
13,456=9,216+7,921-17,088 cos W
13,456=17,137-17,088 cos W
13,456-17,137=17,137-17,088 cos W-17,137
-3,681=-17,088 cos W
(-3,681)/(-17,088)=(-17,088 cos W)/(-17,088)
0.215414326=cos W
cos W = 0.215414326

Solving for W:
W= cos^(-1) 0.215414326
Using the calculator:
W=77.56016397°
Rounded to one decimal place:
W=77.6°

Answer: Third option 77.6°
3 0
4 years ago
There are 15 students in the class and 8 are boys. What is the ratio of girls to students in the room?
cricket20 [7]

Answer:

7/15

Step-by-step explanation:

hope this helps?:)

8 0
3 years ago
Find the roots of h(t) = (139kt)^2 − 69t + 80
Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

7 0
3 years ago
Please may someone help with c
german

Step-by-step explanation:

Distance between two villages = d

(a) \:  \frac{4}{5}  \: of \: d =  \frac{4}{5} d \\  \\ (b) \:  \frac{3}{4}  \: of \: d =  \frac{3}{4} d  \\  \\ (c) \: \frac{4}{5} d -   \frac{3}{4} d = 1.5 \\  \\  \therefore \:   \bigg(\frac{4}{5} -   \frac{3}{4} \bigg) d = 1.5 \\  \\ \therefore \:   \bigg(\frac{4 \times 4 - 3 \times 5}{5 \times 4} \bigg) d = 1.5 \\  \\ \therefore \:   \bigg(\frac{16- 15}{20} \bigg) d = 1.5 \\  \\ \therefore \:   \bigg(\frac{1}{20} \bigg) d = 1.5 \\  \\ \therefore \: d = 1.5 \times 20 \\  \\ \:  \:  \:  \:  \:  \huge \orange{ \boxed{{ \therefore \: d = 30}}}

4 0
4 years ago
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