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goblinko [34]
3 years ago
8

If m<ABF=(7x+20)°,m<FBC=(2x-5)°,and m<ABC=159°, find the value of x​

Mathematics
1 answer:
Bond [772]3 years ago
5 0

Answer:

x = 16

Step-by-step explanation:

∠ ABF + ∠ FBC = ∠ ABC , substitute values

7x + 20 + 2x - 5 = 159 , that is

9x + 15 = 159 ( subtract 15 from both sides )

9x = 144 ( divide both sides by 9 )

x = 16

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I need help with this
olga2289 [7]

Answer:

x = 24.52

Step-by-step explanation:

Since this is a right triangle, use Pythagorean Theorem to solve for the hypotenuse.

Pythagorean Theorem: a² + b² = c²

24² + 5² = x²

576 + 25 = x²

601 = x²

√601 = x

4 0
3 years ago
Parallelogram ABCD has the angle measures shown. Can you conclude that it is a rhombus, a rectangle, or a square? Explain.
Hoochie [10]

Answer:

The correct option is parallelogram ABCD is a rhombus, because the diagonal bisects two angles

Step-by-step explanation:

In triangle ABD:

∠B = ∠D

Thus AB=AC by the property of opposite sides of equal angles are equal

In triangle CBD

∠B = ∠D

Thus CB=CD  by the property of opposite sides of equal angles are equal

Thus all four sides of quadrilateral ABCD are equal  

And diagonal BD bisects the angles  

So, it is a rhombus

Therefore the correct option is  parallelogram ABCD is a rhombus, because the diagonal bisects two angles....

5 0
4 years ago
Helppp me plsssssssss<br><br>​
Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

  • \sf \:\:\:\:\:\:\:\:\:\:x_{k}=35
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k}=50
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42
  • \sf \:\:\:\:\:\:\:\:\:\:h=5

{\bf \:\: {By\:using\:the\: formula}} \\ \\

\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\

\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\

\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\

{\large{\frak{\pmb{\underline{Additional\: information }}}}}

MODE

  • Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

  • In a frequency distribution the class having maximum frequency is called the modal class.

{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\

{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\

Where,

\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.

\small \blue{ \bigstar}\sf \: f_{k}=frequency\:of\:the\:modal\:class

\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class

\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class

\small \purple{ \bigstar}\sf \: h= width \:of\:the\:class\:interval

7 0
3 years ago
Can someone help me with the ones that aren’t done
garri49 [273]

5:

C = 5/9(F - 32)

C = 5/9(F) - 17.777

-5/9(F) + C = -17.777

-5/9(F) = -C - 17.777

F = 5/9(C) + 32.0306

7 0
3 years ago
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stiv31 [10]

Answer:

ygyvyvububububububububububububububububububububububububububububububububububububinininininininininubububyvyctctcrxrxrcrct vvtvybububunininininuninububububububbuub

Step-by-step explanation:

778885542123456678900098776554331112345678900

5 0
3 years ago
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