All categories

2 years ago

What is m∠LQK ? Enter your answer in the box.

Mathematics

2 answers:

Lady_Fox [76]2 years ago

5 0

∠LQK+∠GQL=90º
Therefore:
(4n-15)+(3n)=90
7n=90+15
7n=105
n=105/7
n=15

∠LQK=4n-15=4(15)-15=60-15=45

Answer: ∠LQK=45º

KiRa [710]2 years ago

3 0

Answer:

The measure of ∠LQK is 45° .

Step-by-step explanation:

As the figure is given in the question.

∠LQK + ∠LQG = 90°

As given

∠LQK = 4n - 15

∠LQG = 3n

Putting the values in the above

4n - 15 + 3n = 90

7n - 15 = 90

7n = 15 + 90

7n = 105

$n = \frac{105}{7}$

n = 15

Thus

∠LQK = 4×15 - 15

= 60 - 15

= 45 °

Therefore the measure of ∠LQK is 45° .

You might be interested in

What is the difference in length between the longest and shortest pieces of ribbon?

jasenka [17]

Answer: B

Step-by-step explanation: have a great day pls brainliest

6 0

2 years ago

Read 2 more answers

Sheryl spends $ 12 every week on cigarettes and spends $ 23 a month on dry cleaning. On an annual basis, the money spent on c

Lera25 [3.4K]

Answer:

just for the points

Step-by-step explanation:

6 0

2 years ago

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

2 years ago

Coupon A: off 30% of $63 shoes

Soloha48 [4]

??? huh... i’m so confused pls

3 0

3 years ago

Read 2 more answers

for a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height,

kolbaska11 [484]

Problem

For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground

Solution

We know that the x coordinate of a quadratic function is given by:

Vx= -b/2a

And the y coordinate correspond to the maximum value of y.

Then the best options are C and D but the best option is:

D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a

The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.

7 0

6 months ago