What’s is 3-(-5) picture listed ‍♀️

Mathematics

2 answers:

vichka [17]2 years ago

5 0

The two subtracting signs cancel eachother out to get addition, so the answer is 8.

Pachacha [2.7K]2 years ago

4 0

-15 I believe because u are multiplying negatives

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Biologists have observed that the chirping rate of crickets of a certain species is related to temperature, and the relationship

sveticcg [70]

(a) Find the linear equation that relates the temperature t and the number of chirps per minute n.
If the equation is linear, then the equation should have the form of t = mn + b

m = (80 - 70) / (168 - 120) = 5/24

t = 5/24n + b

70 = 5/24(120) + b
b = 45

t = 5/24 n + 45

(b) If the crickets are chirping at 138 chirps per minute, estimate the temperature. (Round your answer to the nearest degree.)
t = 5/24 n + 45
t = 5/24 (138) + 45 = 74 °F

3 0

3 years ago

Hello please help thanks

Ilya [14]

My educational guess: Most likely Condensation, due to the fact that the in order for precipitation to happen (condensation must occur in short amount of time). Please know this is an educational guess and it may be wrong, you are risking your answer. And if you got it wrong because of me, I’m terribly sorry. God bless you and I hope this helped.

7 0

2 years ago

1. Find the cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of ₹ 3 per m

Flauer [41]

1. Find the cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of ₹ 3 per m².

a = 51 m, b = 37 m, c = 20 m

semiperimeter: p = (51+37+20):2 = 54 m

Area of triangle:

$A=\sqrt{p(p-a)(p-b)(p-c)}\\\\A=\sqrt{54(54-51)(54-37)(54-20)}\\\\A=\sqrt{54\cdot3\cdot17\cdot34}\\\\A=\sqrt{9\cdot2\cdot3\cdot3\cdot17\cdot17\cdot2}\\\\A=3\cdot2\cdot3\cdot17\\\\A=306\,m^2$

Rate: ₹ 3 per m².

Cost: ₹ 3•306 = ₹ 918

2. Find the area of the isosceles triangle whose perimeter is 11 cm and the base is 5 cm.

a = 5 cm

a+2b = 11 cm ⇒ 2b = 6 cm ⇒ b = 3 cm

p = 11:2 = 5.5

$A=\sqrt{5.5(5.5-3)^2(5.5-5)}\\\\ A=\sqrt{5.5\cdot(2.5)^2\cdot0.5}\\\\ A=\sqrt{11\cdot0.5\cdot(2.5)^2\cdot0.5}\\\\A=0.5\cdot2.5\cdot\sqrt{11}\\\\A=1.25\sqrt{11}\,cm^2\approx4.146\,cm^2$

3. Find the area of the equilateral triangle whose each side is 8 cm.

a = b = c = 8 cm

p = (8•3):2 = 12 cm

$A=\sqrt{12(12-8)^3}\\\\ A=\sqrt{12\cdot4^3}\\\\ A=\sqrt{3\cdot4\cdot4\cdot4^2}\\\\A=4\cdot4\cdot\sqrt{3}\\\\A=16\sqrt3\ cm^2\approx27.713\ cm^2$

4. The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle.

a = 2x

b = 3x

2x + 2•3x = 32 cm ⇒ 8x = 32 cm ⇒ x = 4 cm ⇒ a = 8 cm, b = 12 cm

p = 32:2 = 16 cm

$A=\sqrt{16(16-8)(16-12)^2}\\\\ A=\sqrt{16\cdot8\cdot4^2}\\\\ A=\sqrt{2\cdot8\cdot8\cdot4^2}\\\\ A=8\cdot4\cdot\sqrt2\\\\ A=32\sqrt2\ cm^2\approx45.2548\ cm^2$