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3 years ago

PLEASE SOMEONE PLEASE HELP ME PLEASE

Mathematics

1 answer:

kolezko [41]3 years ago

8 0

Answer:

can't see the picture clearly

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2 years ago

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Which expression is equivalent to b^m•b^n<br> A. b^m-n<br> B. b^m•n<br> C. b^m÷n<br> D. b^m+n

Step2247 [10]

$\displaystyle\\ 
\bold{b^m \times b^n = b^{m + n}~~~(D)}\\\\ 
\bold{\frac{b^m}{b^n}= b^{m - n}~~~~~~~~~(A)}\\\\ 
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6 0

3 years ago

please help :{ brainliest if you can get it right and I need you to prove why your answer is correct, thanksss

beks73 [17]

Answer:

The Value of $a=\frac{15}{4}$ .

Step-by-step explanation:

We have Named the figure please find the attachment for your reference.

Given:

PR = y

QR = a

RS = b

PS = z

PQ = x

QS = 15

∠P = 90°

∠R = 90°

∠Q = 60°

∠S = 30°

We need to find the Value of 'a'.

Solution:

Now we know that:

In Δ PQS

∠P = 90°

∠S = 30°

Now we know that;

$sin\ \theta = \frac{opposite\ side}{Hypotenuse}$

$sin \ S= \frac{PQ}{QS}$

Substituting the given values we get;

$sin\ 30\°=\frac{x}{15}$

Now we know that;

$sin\ 30\° = \frac12$

So we can say that;

$\frac{1}{2}=\frac{x}{15}\\\\x=\frac{15}{2}$

Now In Triangle PQR.

∠R = 90°

∠Q = 60°

So we can say that;

$Cos \theta = \frac{adjacent \ Side}{Hypotenuse}\\$

$Cos\ Q = \frac{QR}{PQ}$

Substituting the given values we get;

$cos 60\°= \frac{a}{x}$

Now we know that;

$cos 60\°= \frac12$

$x=\frac{15}{2}$

So substituting the values we get;

$\frac{1}{2}=\frac{a}{\frac{15}{2}}$

By Using Cross Multiplication we get;

$a= \frac{1}{2}\times\frac{15}{2}\\\\a=\frac{15}{4}$

Hence The Value of $a=\frac{15}{4}$ .

6 0

3 years ago