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Y_Kistochka [10]
3 years ago
12

A fourth of the product of 13 and a number M

Mathematics
1 answer:
katrin [286]3 years ago
6 0

Answer:

\frac{13m}{4}

Step-by-step explanation:

"the product of 13 and a number M" should be written as 13m.

To get a fourth of the produce we would need to divide it by four.

A fourth of the product of 13 and a number M should be \frac{13m}{4} .

Hope this helps.

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What is the value of x?​
myrzilka [38]

Answer:

<h2>x = 20</h2><h2 />

Step-by-step explanation:

∠VRW and ∠TRS form a pair of vertical angles then they are congruent

therefore

3x = x + 40

⇔

2x = 40

⇔

x = 20

5 0
3 years ago
a class has ten girls and three boys. if the teacher randomly picks ten students,what is the probability that he will pick all g
grin007 [14]

Answer:

10/13

Step-by-step explanation:

You will first add all the students the mentioned which gives you 13 then since they ask what is the probability it the teacher would pick girls. You know there's 10 girls and 3 boys so it would be 10/13.                                  

7 0
3 years ago
9. In a rhombus whose side measures 34 and the smaller angle is 42°. Find the length of the larger diagonal, to the nearest tent
Veseljchak [2.6K]

Answer:

63.5

Step-by-step explanation:

The diagonals of a rhombus bisect opposite angles.

The longer diagonal bisects the 42° angle

The larger angle of the rhombus is 180 - 42 = 138

Let x = the length of the longer diagonal

You know two sides and the included angle of a triangle.

Using the law of cosines we get

x^{2} = 34^{2} + 34^{2} -2(34)(34) cos 138\\         = 1156 + 1156 - 2312 cos 138\\         = 4030.150837...\\x = \sqrt{4030.150837} = 63.48x^{2} = 34^{2} + 34^{2} -2(34)(34)cos 138\\         = 1156 + 1156 - 2312 cos 138\\         = 4030.1508...\\     x = \sqrt{4030.1508} \\          =  63.48x^2 = 34^2 + 34^2 - 2(34)(34) cos 138

      = 1156 + 1156 - 2312 c0s 138

      = 4030.1508...

x = sq rt 4030.1508...

  =  63.5

7 0
3 years ago
X^3 + ax^2 +2ax + 5 divided by x + 1 what is the value of a
svlad2 [7]

a

X

3

x

+

1

+

a

2

x

2

x

+

1

+

2

a

2

x

x

+

1

+

5

a

x

+

1

5 0
3 years ago
At a certain time of the day, a tree that is x meters tall casts a shadow that is x-49 meters long. If the distance from the top
nikklg [1K]

The height of the tree is 60 meters.

Explanation:

Let the height of the tree be x. The tree casts a shadow of x-49 meters and the distance from the top of the tree to the end of the shadow is x+1 meters.

The sides of the triangle are attached in the image below:

Using pythagoras theorem,

x^{2}+(x-49)^{2}=(x+1)^{2}

Expanding, we get,

2 x^{2}-98 x+2401=x^{2}+2 x+1

2 x^{2}-98 x+2400=x^{2}+2 x

2 x^{2}-100 x+2400=x^{2}

x^{2}-100 x+2400=0

Solving the equation using the quadratic formula x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}, we get,

x=\frac{-(-100)\pm\sqrt{(-100)^{2}-4 \cdot 1 \cdot 2400}}{2 \cdot 1}

Simplifying, we have,

x=\frac{100\pm\sqrt{10000-9600}}{2}

x=\frac{100\pm\sqrt{400}}{2}

x=\frac{100\pm{20}}{2}

Thus,

x=\frac{100+20}{2} \\x=\frac{120}{2} \\x=60  and  x=\frac{100-20}{2} \\x=\frac{80}{2} \\x=40

where the value x=40 is not possible because substituting the value x=40 in x-49 results in negative solution. Which is not possible.

Hence, the value of x is 60.

Thus, The height of the tree is 60 meters.

7 0
3 years ago
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