All categories

2 years ago

If ON = OL find mOML

Mathematics

2 answers:

Zanzabum2 years ago

4 0

Answer:

The correct answer is option A. 26°

Step-by-step explanation:

From the figure we can see that,

ON = OL

OM ⊥ NL

Therefore m<N = m<L

<u>To find the value of x</u>

m<N = m<L

7x + 8 = 9x -8

9x - 7x = 8 + 8

2x = 16

x = 8

<u> To find the value of m<OML</u>

x = 8

m<L = 9x - 8

= 9*8 - 8 = 72 - 8 = 64

m<OML = 180 - (<L + <O)

= 180 - ( 64 + 90)

= 26°

Therefore the correct answer is option A. 26°

saveliy_v [14]2 years ago

3 0

Answer:

A. 26°

Step-by-step explanation:

If ON =OL, then the two triangles are are similar.

The base angles must be equal.

$(9x-8)\degree=(7x+8)\degree$

We group the similar terms to obtain:

$9x-7x=8+8$

$2x=16$

Divide through by 2 to get:

x=8

$m\angle OML=90-(7x+8)$

Substitute x=8

$m\angle OML=90-(7(8)+8)$

$m\angle OML=90-(56+8)$

$m\angle OML=26\degree$

You might be interested in

Does anyone know this

Tamiku [17]

Answer:

I belive the answer is A

Step-by-step explanation:

So any answer with 22t would make sense, so you have A and C. In C though, it is subtracting 22, but since 6195 is the total it would have to include the 22 so it is A.

8 0

2 years ago

The figure below shows square ABCD inscribed on square PQRM on a coordinate plane. Which of the following expressions represents

vova2212 [387]

The hypotenuse of the triangle is given by:
$CB ^ 2 = RC ^ 2 + RB ^ 2
$
Substituting values:
$CB ^ 2 = 2 ^ 2 + 4 ^ 2
$
Clearing CB we have:
$CB = \sqrt{ 2 ^ 2 + 4 ^ 2} 

 CB = \sqrt{ 4 + 16} 

 CB = \sqrt{20} 

 CB = \sqrt{4*5} 

 CB = 2\sqrt{5} 



$
Then, the perimeter of the triangle is:
$P = RB + RC + CB
$
Substituting values:
$P = 4 + 2 + 2 \sqrt{5} 

 P = 6 + 2 \sqrt{5}$
Answer:
the perimeter of triangle BRC in units is:
D.6 + 2√5

7 0

2 years ago

Read 2 more answers

The half-life of cesium-137 is 30 years. Suppose we have a

VMariaS [17]

(a) If $y(t)$ is the mass (in mg) remaining after $t$ years, then $y(t) = y(0) e^{kt}=100e^{kt}.$

$y(30) = 100 e^{30k} = \frac{1}{2}(100) \implies e^{30k} = \frac{1}{2} \implies k = -(\ln 2) /30 \implies \\ \\
y(t) = 100e^{-(\ln 2)t/30} = 100 \cdot 2^{-t/30}$

(b) $y(100) = 100 \cdot 2^{-100/30} \approx \text{9.92 mg}$

(c)
$100 e^{- (\ln 2)t/30} = 1\ \implies\ -(\ln 2) t / 30 = \ln \frac{1}{100}\ \implies\ \\ \\
t = -30 \frac{\ln 0.01}{\ln 2} \approx \text{199.3 years}$

3 0

2 years ago

Helpppppppppppppppp ill mark you brainlist write in y=mx+b form

Sholpan [36]

Answer:

The equation of the line in slope-intercept form is: