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3 years ago

7

A rectangular barn door is being constructed using its diagonals, as illustrated in the image below. the doorknob is placed at t

he intersection of the 2 diagonals. what is the length of x if each diagonal measures 10 feet?
A. 20ft
B. 15ft
C. 5ft
D. 10ft

2 answers:

eduard3 years ago

4 0

The answer is letter c, five feet

Lubov Fominskaja [6]3 years ago

3 0

Answer:

Correct option is:

C. 5ft

Step-by-step explanation:

Square or rectangle or rhombus or parallelogram: In all these four shapes, diagonals bisect each other.

Here we are given a rectangle whose each diagonal measures 10 feet

diagonal bisect each other

So, the length of x=5 feet

Hence, Correct option is:

C. 5ft

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Ierofanga [76]

Answer: 2.73 cm

Step-by-step explanation:

Given that :

Side (a) of cube = 4.4cm

Volume of a cube (V) = a^3

V = 4.4^3

V = 85.184cm^3

Therefore, volume of the sphere made = 85.184cm^3

Volume of sphere = 4/3 πr^3

Where r = radius

85.184 = (4/3)*(22/7)*r^3

85.184 = (88/21)*r^3

85.184 = 4.1905 * r^3

r^3 = 85.184 / 4.1905

r^3 = 20.327884

Take the cube root of both sides

r = 2.73 cm

5 0

3 years ago

Solve for AB<br><br><br> a: 3.6<br> b: 15<br> c: 4<br> d: 10

IrinaK [193]

A because 9 divided by 15 equals 0.6. Then I multiply 0.6 times 6 which equals 3.6. I didn’t 9 divided by 15 because they are congruent.

4 0

3 years ago

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

4 years ago

Help!!! This question is confusing me!!

erica [24]

Answer:

k ≥2

Step-by-step explanation:

k+ 1/3≥7/3

Subtract 1/3 from each side

k+ 1/3-1/3≥7/3-1/3

k ≥6/3

k ≥2

8 0

3 years ago

mihalych1998 [28]

Answer:

$=4.+2.$

Step-by-step explanation:

<u>Linear Combination Of Vectors </u>

One vector $\vec b$ is a linear combination of $\vec a_1$ and $\vec a_2$ if there are two scalars $x_1, x_2$ such as

$\vec b=x_1\vec a_1+x_2\vec a_2$

In our case, all the vectors are given in $R^3$ but there are only two possible components for the linear combination. This indicates that only two conditions can be used to determine both scalars, and the other condition must be satisfied once the scalars are found.

We have

$\vec a_1=,\ \vec a2=,\ \vec b=$

We set the equation

$=x_1.+x_2.$

Multiplying both scalars by the vectors

$=+$

Equating each coordinate, we get

$4x_1-4x_2=8$

$5x_1+3x_2=26$

$-4x_1+3x_2=-10$

Adding the first and the third equations:

$-x_2=-2$

$x_2=2$

Replacing in the first equation

$4x_1-4(2)=8$

$4x_1=8+8$

$x_1=4$

We must test if those values make the second equation become an identity

$5(4)+3(2)=20+6=26$

The second equation complies with the values of $x_1$ and $x_2$ , so the solution is

$=4.+2.$

8 0

3 years ago