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3 years ago

Solve for d: K + 7d = 6c − 10d.

Mathematics

2 answers:

evablogger [386]3 years ago

8 0

Answer:

The answer to your question is d = (6c - K)/17

Step-by-step explanation:

Data

K + 7d = 6c - 10d

Process

1.- Add 10d in both sides

K + 7d + 10d = 6c - 10d + 10d

2.- Simplify

K + 17d = 6c

3.- Subtract K in both sides

K + 17d - K = 6c - K

4.- Simplify

17d = 6c - K

5.- Divide both sides by 17

17d/17 = (6c - K)/17

6.- Simplify

d = (6c - K)/17

Wewaii [24]3 years ago

5 0

Answer:

$d = \frac{6c}{17} - \frac{k}{17}$

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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sue had 1 million dollars and ben had 1 dollars how much more did sue had after they spent 6 dollars out of there account

Naddika [18.5K]

Ben had -5 dollars, and sue had nine hundred ninety-nine thousand nine hundred ninety-four dollars.

3 0

3 years ago

Round each factor to the nearest ten to estimate the product of 34 . 39

aev [14]

Answer:

Round 34 down to 30 then round 39 up to 40

Step-by-step explanation:

34 ⟶ 30 34 is rounded down to 30

39 ⟶ 40 39 rounded up to 40

34 ⟶ 30

39 ⟶ 40

34 is rounded down to 30

39 rounded up to 40

Calculate mentally 34 × 39 = 1326

The estimated product is 1326.

3 0

3 years ago

Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137

Nata [24]

Answer:

<em>Camera 2nd has to cover the maximum angle, i.e. </em> $78.70^\circ$ .

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle $\triangle ABC$ with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles $\angle A, \angle B, \angle C$

As per Law of cosine:

$cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}$

Putting the values of a,b and c to find out angles $\angle A, \angle B, \angle C$ .

$cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ$

$cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ$

$cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ$

<em>Camera 2nd has to cover the maximum angle</em>, i.e. $78.70^\circ$ .

6 0

4 years ago

An airplane takes off from an airport and travels 1100 miles east and then travels 600 miles north. This trip can be plotted on

IgorLugansk [536]

Answer:

The plane is now at (1100, 600) in unit mile

Step-by-step explanation:

The airport is the point (0 ,0) and the plane travels from there 1100 miles east and thereafter 600 miles north. If the unit is in mile then we may say that the planes co -ordinates are now (1100 , 600)

Since,

positive x -axis points towards east and positive y-axis points towards north.

4 0

3 years ago

20r3 − 16r2 + 15r − 12

kati45 [8]

Answer:

(5r-4)(4r^(2)+3)

Step-by-step explanation:

Hope it helps

7 0

3 years ago