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Andrew [12]
3 years ago
10

5 is the same as ___ and ___.

Mathematics
1 answer:
nirvana33 [79]3 years ago
4 0
1/2 and 1/3 1/4. Is the answer I automatically thought of
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5 Find the slope of the line that contains points (1,-3) and (-4,5).
ch4aika [34]

Answer:

gradient = m

    m=\frac{rise}{run}

m=\frac{1-(-4)}{-3-5}

m=\frac{5}{-8}

3 0
3 years ago
What is the measure of ∠P?
Kobotan [32]

Answer:

65°

Step-by-step explanation:

all angles in a trapezium equate to 360°

R is 115°

115+115=230

360-230=130

130÷2=65

8 0
3 years ago
Read 2 more answers
Find the solution of the differential equation dy/dt = ky, k a constant, that satisfies the given conditions. y(0) = 50, y(5) =
irga5000 [103]

Answer:  The required solution is y=50e^{0.1386t}.

Step-by-step explanation:

We are given to solve the following differential equation :

\dfrac{dy}{dt}=ky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.

From equation (i), we have

\dfrac{dy}{y}=kdt.

Integrating both sides, we get

\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]

Also, the conditions are

y(0)=50\\\\\Rightarrow ae^0=50\\\\\Rightarrow a=50

and

y(5)=100\\\\\Rightarrow 50e^{5k}=100\\\\\Rightarrow e^{5k}=2\\\\\Rightarrow 5k=\log_e2\\\\\Rightarrow 5k=0.6931\\\\\Rightarrow k=0.1386.

Thus, the required solution is y=50e^{0.1386t}.

8 0
3 years ago
Read 2 more answers
The coordinates of the endpoints of AB and CD are A(2,
Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

m= \frac{y_2-y_1}{x_2-x_1}

Formula to Find Distance between two points

D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

mAB ( represents , Slope of AB )

1. mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}

2. mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}

3. mCD=\frac{5-2}{6-5}=\frac{3}{1}=3

4. AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}

5. BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}

mAC = mBC  , and C is common point , hence these three are collinear points  making a straight line whole slope is -\frac{1}{3}

mAB=-\frac{1}{3}

mCD=3

mAB \times mCD = -\frac{1}{3} \times 3 = -1

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0
3 years ago
If y=7 when x=64 what is x when y=8
kirill [66]

Answer:

when y=8 x= 73 ❤️❤️❤️❤️❤️

3 0
3 years ago
Read 2 more answers
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