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3 years ago

In triangle JKL, solve for x

Mathematics

2 answers:

Archy [21]3 years ago

8 0

For this question you need to apply the trigonometric ratios with cos 27 degrees
cos 27 = $\frac{34}{x}$
0.891x = 34cm
∴x = 38.159cm

please mark as brainliest

Alchen [17]3 years ago

3 0

38.16 = when rounded

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masha68 [24]

Are you looking for c? if so c=5

6 0

2 years ago

Read 2 more answers

What is the initial value of y= 8x+7

allochka39001 [22]

Answer:

Step-by-step explanation:

The initial value is another name for the y-intercept. This equation is written in point slope form (y=mx+b). And before you think y-intercept means y, right? No that is incorrect, I got that wrong in the beginning too ;)

y=mx+b

y=8x+7

b in the point slope form equation is always the initial value or y-intercept or whatever people call it. So what substitutes b out here? 7. So 7 is the initial value!

:) Hope you understand now! Have a good day!

8 0

2 years ago

1/2b+9/4=7/4 solve the equation

AlladinOne [14]

First subtract 9/4 from each side

1/2b=-2/4

Simplify the 2/4

1/2b=-1/2

Divide each side by 1/2

b = -1

4 0

3 years ago

Which ordered pair is a solution of the equation?<br> Y-4=7(x-6)

deff fn [24]

Answer:

(6,4)

Step-by-step explanation:

The equation is in 'point-slope' form.

$y-y_1=m(x-x_1)\\\rule{150}{0.5}\\(x_1,y_1) \rightarrow \text{a point that's given}\\\\m \rightarrow \text{slope}$

This means we can identify the point that was used in the equation.

$y-\boxed{y_1} =m(x-\boxed{x_1}) | y-\boxed{4}=7(x-\boxed{6})\\\rule{210}{0.5}\\y_1 = 4\\\\x_1 =6\\\rule{210}{0.5}\\(x_1,y_1) \rightarrow \boxed{(6,4)}$

(6,4) would be a solution to the equation given.

Hope this helps.

4 0

2 years ago

Simplify the following Questions. (IMPORTANT NOTE!!! #'s 2 AND 7 MUST BE IN SCIENTIFIC NOTATION!!!!!) 1.) (x)²(2xy³)^5 2.) (4x10

poizon [28]

Answer:

Step-by-step explanation:

Hint :

$(a*b)^{m}= a^{m}b^{m}\\\\\\(a^{m})^{n}=a^{mn}\\\\\\a^{m}*a^{n}=a^{m+n}\\$

$1)x^{2}(2xy^{3})^{5}=x^{2}*2^{5}*x^{5}*(y^{3})^{5}\\\\\\=x^{2}*2^{5}*x^{5}*y^{3*5}\\\\=x^{2}*2^{5}*x^{5}*y^{15}\\\\=2^{5}*x^{2+5}*y^{15}\\\\=2^{5}x^{7}y^{15}$

$2) (4x10^{8})^{2}=4^{2}*x^{2}*(10^{8})^{2}\\\\=16*x^{2}*10^{8*2}\\\\=1.6*10*x^{2}*10^{16}\\\\=1.6x^{2}*10^{16+1}\\\\=1.6x^{2}10^{17}$

$3)(-h^{4})^{5}=(-h)^{4*5}=-h^{20}\\\\\4)(3xy^{3})^{2}(xy)^{6}=3^{2}x^{2}(y^{3})^{2}x^{6}y^{6}\\\\=9x^{2}y^{3*2}x^{6}y^{6}\\\\\\=9x^{2}y^{6}x^{6}y^{6}\\\\=9x^{2+6}y^{6+6}\\\\=9x^{8}y{12}\\\\\\$

$5) (p^{9})^{-2}=p^{9*-2}=p^{-18}\\\\\\6)(5k^{2})^{3}=5^{3}(k^{2})^{3}=625k^{6}\\\\7)(7x10^{5})^{2}=7^{2}x^{2}(10^{5})^{2}\\\\=49x^{2}10^{5*2}\\\\=4.9*10^{1}x^{2}10^{10}\\\\=4.9x^{2}10^{10+1}\\\\=4.9x^{2}10^{11}$

$8)x^{3}(-x^{3}y)^{2}=x^{3}*(-x^3})^{2}y^{2}\\\\\\=x^{3}*(-1)^{2}*x^{2*3}y^{2}\\\\=x^{3}*1*x^{6}y^{2}=x^{3+6}y^{2}\\\\=x^{9}y^{2}\\\\\\9)w^{5}(w^{2})^{-4}=w^{5}w^{2*-4}\\\\=w^{5}w^{-8}\\\\=w^{5-8}=w^{-3}\\\\=\frac{1}{w^{3}}\\\\\\10) (2x^{5})^{4}=2^{4}x^{5*4}\\\\=2^{4}x^{20}\\\\=16x^{20}$

5 0

2 years ago