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2 years ago

15

To solve this question you have to use TAN COS SIN i give brainiest

2 answers:

ElenaW [278]2 years ago

7 0

Answer:

4.3

Step-by-step explanation:

<u>Use cosine:</u>

cos = adjacent / hypotenuse
cos ∠L = KL/JL
cos 21° = 4/x
x = 4 / cos 21°
x = 4.3 (rounded)

kati45 [8]2 years ago

3 0

Answer:

4.3

Step-by-step explanation:

Here we will use cosine to find the value..

<h3> $\cos = \frac{adjacent}{hypotenuse}$ </h3><h3> $\cos∠l = \frac{kl}{jl}$ </h3><h3> $\cos21° = \frac{4}{x}$ </h3><h3> $x = \frac{4}{ \cos21° }$ </h3><h3> $x = 4.3$ When rounded off...</h3>

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Based on the table, what is Nick's rate?

Murrr4er [49]

21 /3 = 7
35 / 5 = 7
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2 years ago

Read 2 more answers

"If 5 is decreased from double Joe's weight, the result is 230. What is Joe's weight?"<br> 30

Gala2k [10]

Answer:

117.5 or 235/2

Step-by-step explanation:

We can set up an equation for this problem:

Lets Joe's weight be x

then:

2x -5 = 230

therefore:

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x = 117.5 or 235/2

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3 years ago

After sales tax, an $81 cell phone costs $84.24. What is the sales tax percentage?

Igoryamba

Answer:

4% sales tax

Step-by-step explanation:

Take the original costs and subtract it from the price after sales tax.

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2 years ago

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

6 0

2 years ago

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kumpel [21]

No sé, prueba 2,3,5,7,11,13....
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