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1 year ago

Why are the solutions to the proportions 50/x =10/20 and 10/50=20/x the same

Mathematics

2 answers:

storchak [24]1 year ago

7 0

Answer:

because both result in the equation 10x=1000 which simplifies to x=100.

ratelena [41]1 year ago

5 0

With cross multiplication you can find that they are the same. In both equations, x would be 100.

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Which statements about the hyperbola are true? Check all that apply.

mariarad [96]

Answer:

True options: 1, 2 and 5

Step-by-step explanation:

From the given diagram, you can see that the center of the hyperbola is placed at the origin, so first option is true (see attached diagram for definition of center, vertices, foci, i.e.)

There are two vertices of the hyperbola, they are placed at (-6,0) and (6,0), so second option is true.

The transverse axis is the segment connecting vertices, this segment is horizontal, so option 3 is false.

The foci are not placed within the rectangular reference box, so this option is false.

The directrices are vertical lines with equations $x=\pm \dfrac{a}{e}$ , so this option is true.

8 0

2 years ago

Read 2 more answers

∆ABC is translated 2 units down and 1 unit to the left. Then it is rotated 90° clockwise about the origin to form ∆A′B′C′.

RideAnS [48]

Answer:

The coordinates of image are A'(-2,1), B'(1,0) and C'(-1,0).

Step-by-step explanation:

From the figure it is clear that the coordinates of triangle are A(0,0), B(1,3) and C(1,1).

∆ABC is translated 2 units down and 1 unit to the left.

$(x,y)\rightarrow (x-1,y-2)$

$A(0,0)\rightarrow (0-1,0-2)=(-1,-2)$

$B(1,3)\rightarrow (1-1,3-2)=(0,1)$

$C(1,1)\rightarrow (1-1,1-2)=(0,-1)$

Then it is rotated 90° clockwise about the origin to form ∆A′B′C′.

$(x,y)\rightarrow (y,-x)$

$(-1,-2)\rightarrow A'(-2,1)$

$(0,1)\rightarrow B'(1,0)$

$(0,-1)\rightarrow C'(-1,0)$

Therefore the coordinates of image are A'(-2,1), B'(1,0) and C'(-1,0).

7 0

1 year ago

From home, Mary’s work is two thirds along the way to training. Training is 2.5km from work. Mary normally goes to work, then tr

polet [3.4K]

First thing to do is to illustrate the problem, Since it was mentioned that work was along the way to training, the order is shown in the picture. Mary's home and workplace are nearer compared to her training center. It is also mentioned that the distance between work and home, denoted as x, is 2/3 of the total distance from home to training. The total distance is (x + 2.5). Thus,

x = 2/3(x+2.5)
x = 2/3 x + 5/3
1/3 x = 5/3
x = 5 km

Thus, the distance from home to work is 5 km. This means that Mary has to walk this distance twice to return home to get her shoes. Then, she will travel again the total distance of 5+2.5 = 7.5 km to get to her training center. So,

Total distance = 2(5km) + 7.5 km
Total distance = 17.5 km

3 0

1 year ago

381.10-214.43 equals

dmitriy555 [2]

Answer:

166.67

Step-by-step explanation:

3 0

2 years ago

Please answer all of them need this

VikaD [51]

First Question

For a better understanding of the solution provided here please find the first attached file which has the diagram of the the isosceles trapezoid.

We dropped perpendiculars from C and D to intersect AB at Q and P respectively.

As can be seen in $\Delta BCQ$ , we can easily find the values of CQ and BQ.

Since, $Sin(75^0)=\frac{CQ}{8}$

$\therefore CQ=8\times Sin(75^0)\approx 7.73$ ft

In a similar manner we can find BQ as:

$Cos(75^0)=\frac{BQ}{8}$

$BQ\approx2.07$ ft

All these values can be found in the diagram attached.

Thus, because of the inherent symmetry of the isosceles trapezoid, PQ can be found as:

$PQ=22-(AP+QB)=22-(2.07+2.07)=17.86$

Let us now consider $\Delta AQC$

We can apply the Pythagorean Theorem here to find the length of the diagonal AC which is the hypotenuse of $\Delta AQC$ .

$AC=\sqrt{(AQ)^2+(QC)^2}=\sqrt{(AP+PQ)^2+(QC)^2}=\sqrt{(2.07+17.86)^2+(7.73)^2}\approx21.38$ feet.

Thus, out of the given options, Option B is the closest and hence is the answer.

Second Question

For this question we can directly apply the formula for the area of a triangle using sines which is as:

Area= $\frac{1}{2}$ (First Side)(Second Side)(Sine of the angle between the two sides)

Thus, from the given data,

Area= $\frac{1}{2}\times 218.5\times 224.5\times sin(58.2^0)\approx20845$ $m^2$

Therefore, Option D is the correct option.

Third Question

For this question we will apply the Sine Rule to the $\Delta ABC$ given to us.

Thus, from the triangle we will have:

$\frac{AB}{Sin(\angle C)}=\frac{BC}{Sin(\angle A)}$

$\frac{c}{Sin(\angle C)}=\frac{a}{Sin(\angle A)}$

$\frac{17}{Sin(25^0)}=\frac{a}{Sin(45^0)}$

This gives $a$ to be:

$a\approx28.44$

Which is not close to any of the given options.

Fourth Question

Please find the second attachment for a better understanding of the solution provided her.

As can be clearly seen from the attached diagram, we can apply the Cosine Rule here to find the return distance of the plane which is CA.

$AC=\sqrt{(AB)^2+(BC)^2-2(AB)(BC)\times Cos(\angle B)}$

$\therefore AC=\sqrt{(172.20)^2+(111.64)^2-2(172.20)(111.64)\times Cos(177.29^0)}\approx283.8$ miles.

Thus, Option D is the answer.

8 0

2 years ago

Read 2 more answers