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3 years ago

Calculate the above formula, expressing the answer in scientific notation with the correct number of significant figures

1 answer:

4 0

$\frac{8.86 + 1.0 \times 10^{-3}} {3.610 \times 10^{-3}}=\frac{8.86 + 0.001} {0.00361} \\ \\ =\frac{8.861} {0.00361}=2,454.57=\bold{2.455\times10^3}$

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Solve for y. 15y - -9y + -10y = 14 y = Submit

Tems11 [23]

Answer:

I think y=1

Step-by-step explanation:

15y+9y= 24y

24y-10y=14y

divide 14y by 14 and get just y and divide 14 by 14 to get one

y=1

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3 years ago

Find the distance between (6, 0) and (10, 3).

yulyashka [42]

A 5 units I believe

6 0

2 years ago

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3. If 2p + q = 11 and p + 2q = 13, then p+q= (A) 6 (B) 8 (C) 9 (D) 12 (E) 18

frozen [14]

Answer:
(C) 9

Step by step:
11-2=9 and it’s asking for both p and q combined so no need to worry about that

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3 years ago

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Giving brainliest! Please show the working out

tiny-mole [99]

Answer:

see explanation

Step-by-step explanation:

(a)

OC = OB ( both radii of the circle )

Thus Δ BOC is isosceles with congruent base angles.

∠ BOC = ∠ BCO = 50°

(b)

∠ ACB = 90° ( angle in a semicircle ), then

∠ ACO = 90° - 50° = 40°

OA = OC ( both radii of the circle )

Thus Δ ACO is isosceles with congruent base angles.

∠ BAC = ∠ ACO = 40°

7 0

3 years ago

Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar

Brrunno [24]

The distance between the two points is $d=8.6m$

The polar coordinate of A is $\left(4.47,296.57\right)$

The polar coordinate of B is $\left(4.24,135\right)$

Explanation:

The two points are $A(2,-4)$ and $B(-3,3)$

The distance between two points is given by,

$d=\sqrt{(2+3)^{2}+(-4-3)^{2}}\\d=\sqrt{(5)^{2}+(-7)^{2}}\\d=\sqrt{25+49}\\d=8.6$

Thus, the distance between the two points is $d=8.6m$

The polar coordinates of A can be written as $(Distance, tan^{-1} \frac{y}{x} )$

Distance = $\sqrt{x^{2} +y^{2} }$

Substituting $A(2,-4)$ , we get,

Distance = $\sqrt{2^{2} +(-4)^{2} }=\sqrt{4+16 }=4.47$

$tan^{-1} \frac{y}{x} =tan^{-1} \frac{-4}{2}=-63.43$

To make the angle positive, let us add 360,

$\theta=360-63.43=296.57$

The polar coordinate of A is $\left(4.47,296.57\right)$

Similarly, The polar coordinate of B can be written as $(Distance, tan^{-1} \frac{y}{x} )$

Distance = $\sqrt{x^{2} +y^{2} }$

Substituting $B(-3,3)$ , we get,

Distance = $\sqrt{(3)^{2}+(3)^{2}}=4.24$

$tan^{-1} \frac{y}{x} =tan^{-1} \frac{3}{-3}=-45$

To make the angle positive, let us add 360,

$\theta=180-45=135^{\circ}$

The polar coordinate of B is $\left(4.24,135\right)$

5 0

3 years ago