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2 years ago

Please help me!?))),)$&$(

Mathematics

1 answer:

Solnce55 [7]2 years ago

7 0

Answer:

Put a point at -5 on the Y axis.

Then move up 6 (still on the Y axis)

Then move to the right 1 (this should be the point 1,1 on the graph)

Draw a line through the points

Shade to the right of the line you drew

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Estimate a 15% tip on a dinner bill of $39.42 by first rounding the bill amount to the nearest ten dollars.

Tpy6a [65]

So we first round the bill to the nearest ten dollars which would be : $40.
Now we take $40 x 0.15 = $6.00
The answer is : $6 tip

4 0

3 years ago

Find the value of 7!?

Schach [20]

The question isn’t clear, post the question so I could help :) !

5 0

3 years ago

Read 2 more answers

I need help with questions #7 and #8 plz

katen-ka-za [31]

Answer:

7. A = 40.8 deg; B = 60.6 deg; C = 78.6 deg

8. A = 20.7 deg; B = 127.2 deg; C = 32.1 deg

Step-by-step explanation:

Law of Cosines

$c^2 = a^2 + b^2 - 2ab \cos C$

You know the lengths of the sides, so you know a, b, and c. You can use the law of cosines to find C, the measure of angle C.

Then you can use the law of cosines again for each of the other angles. An easier way to solve for angles A and B is, after solving for C with the law of cosines, solve for either A or B with the law of sines and solve for the last angle by the fact that the sum of the measures of the angles of a triangle is 180 deg.

We use the law of cosines to find C.

$18^2 = 12^2 + 16^2 - 2(12)(16) \cos C$

$324 = 144 + 256 - 384 \cos C$

$-384 \cos C = -76$

$\cos C = 0.2$

$C = \cos^{-1} 0.2$

$C = 78.6^\circ$

Now we use the law of sines to find angle A.

Law of Sines

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

We know c and C. We can solve for a.

$\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$

$\dfrac{12}{\sin A} = \dfrac{18}{\sin 78.6^\circ}$

Cross multiply.

$18 \sin A = 12 \sin 78.6^\circ$

$\sin A = \dfrac{12 \sin 78.6^\circ}{18}$

$\sin A = 0.6535$

$A = \sin^{-1} 0.6535$

$A = 40.8^\circ$

To find B, we use

m<A + m<B + m<C = 180

40.8 + m<B + 78.6 = 180

m<B = 60.6 deg

I'll use the law of cosines 3 times here to solve for all the angles.

Law of Cosines

$a^2 = b^2 + c^2 - 2bc \cos A$

$b^2 = a^2 + c^2 - 2ac \cos B$

$c^2 = a^2 + b^2 - 2ab \cos C$

Find angle A:

$a^2 = b^2 + c^2 - 2bc \cos A$

$8^2 = 18^2 + 12^2 - 2(18)(12) \cos A$

$64 = 468 - 432 \cos A$

$\cos A = 0.9352$

$A = 20.7^\circ$

Find angle B:

$b^2 = a^2 + c^2 - 2ac \cos B$

$18^2 = 8^2 + 12^2 - 2(8)(12) \cos B$

$324 = 208 - 192 \cos A$

$\cos B = -0.6042$

$B = 127.2^\circ$

Find angle C:

$c^2 = a^2 + b^2 - 2ab \cos C$

$12^2 = 8^2 + 18^2 - 2(8)(18) \cos B$

$144 = 388 - 288 \cos A$

$\cos C = 0.8472$

$C = 32.1^\circ$

8 0

3 years ago

Greta added 3/6 of a cup of sugar to her cake mix, then sprinkled

bazaltina [42]

3/6 of a cup, 1/6 times 3/1 equals 3/6 or 1/2 :)

8 0

3 years ago

A thermometer is taken from a room where the temperature is 24oc to the outdoors, where the temperature is −15oc. After one minu

kap26 [50]

Solution:

Use Newton's Law of Cooling.

T = T_s + (T_0 - T_s)*e^(-kt)

where

T = temperature at any instant

T_s = temperature of surroundings

T_0 = original temperature

t = elapsed time

k = constant

Now, we need to find this constant. We are given that after one hour, the temperature drops to 13° C in a 7°C Environment.

T = 14, T_0 = 24, T_s = -15, t = 1, k = ?

T = T_s + (T_0 - T_s)*e^(-kt)

==> 14 = -15 + (24 - 7)*e^(-k)

==> 14 = 7 + 17*e^(-k)

==> 7 = 17*e^(-k)

==> 7/13 = e^(-k)

==> -k = ln(7/17)

==> k = -ln(7/17) ≈ 0.774

Now,

Let's calculate temperatures!

T = ?, T_0 = 24, T_s = -15, k = 0.773, t = 3

T = T_s + (T_0 - T_s)*e^(-kt)

==> T = -15 + (24 –(-15))*e^[ -(0.774)(2) ]

==> T = -15 + 39*e^(-1.548)

==> T ≈ 15.72° C

This the required answer.

7 0

3 years ago