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2 years ago

I dont understand this

Mathematics

1 answer:

MrRa [10]2 years ago

8 0

Answer: First option.

Step-by-step explanation:

You need to remember the following Trigonometric Identity:

$cos\alpha =\frac{adjacent}{hypotenuse}$

In this case, you can identify in the figure that:

$\alpha =35\°\\adjacent=9\\hypotenuse=x$

Then you must substitute these values into $cos\alpha =\frac{adjacent}{hypotenuse}$ :

$cos(35\°) =\frac{9}{x}$

Finally, you need to solve for "x" in order to find its value. This is:

$(x)(cos(35\°))=9\\\\x=\frac{9}{cos(35\°)}\\\\x=10.98$

Rounding the result to the nearest tenth, you get:

$x\approx11$

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a rectangle has sides that are 23 m 19 m and 19 m what is a rectangles lengthA. 19 mB. 23 mC. 84 mD. 46 mthis is what it said Th

sdas [7]

Given Data:

The sides of the rectangle are 23 m, 23 m ,19 m and 19 m.

The length of the rectangle is 23 m.

Thus, option (B) is the correct option.

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11 months ago

Ben starts walking along a path at 3 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the sam

Travka [436]

Answer:

2 hours 15 minutes

Step-by-step explanation:

Given that :

Ben's speed = 3 mi/hr

One and half hours after Ben leaves = 1.5 hours

Ben's distance = 3 mi/hr * 1.5 = 4.5 miles

Amanda's speed = 5 mi/hr

Time taken before Amanda catches up:

Let time taken = x

Distance = speed * time

Ben's distance = Amanda's distance

4.5 mi + (3 * x) = 5 * x

4.5 + 3x = 5x

4.5 = 5x - 3x

4.5 = 2x

x = 4.5 / 2

x = 2.25 hours

Time taken = 2.25 * 60 = 135 minutes = 2 hours 15 minutes

7 0

2 years ago

I'm struggling in Algebra 2 right now! Can someone please explain how to work out

IRINA_888 [86]

First of all, we *do* have to put explanations, so it's good that you want them, keep this attitude going!

As for the excercise, we want to use the property that allows us to swap sum/subtractions of logarithm into multiplication/division inside them:

$\log(a)+\log(b) = \log(ab),\quad \log(a)-\log(b)=\log\left(\dfrac{a}{b}\right)$

So, we want to move all logarithmic terms to the left hand side: add $\log_2(2-x)$ to both sides to get

$\log_2(-x)+\log_2(2-x) = 3$

Now we can use the property we just invoked:

$\log_2(-x(2-x)) = 3$

This is a good position, because every equation of the form

$\log_a(b)=c$

can be solved by considering both sides as the exponents of the base of the logarithm:

$\log_a(b)=c \iff a^{\log_a(b)}=a^c \iff b = a^c$

(I used the fact that exponential and logarithm are the inverse one of the other: $a^{\log_a(b)}=b$ )

So, we can conclude

$-x(2-x) = 2^3 \iff -2x+x^2 = 8 \iff x^2-2x-8=0$

This is a standard quadratic equation, whose solutions are $x=-2,\ x=4$ (I assume you are confident in solving this kind of equation, let me know otherwise)

Finally, we have to check if these solutions can actually be fed in the equation: we have

$x=-2 \implies \log_2(2)=3-\log_2(2+2) \iff 1=3-2$

which is true. On the other hand,

$x=4 \implies \log_2(-4)=3-\log_2(2-4)$

But this expression can't be computed, because you can't compute the logarithm of a negative number. So, the only feasible solution is $x=-2$

6 0

3 years ago

Please please help ASAP I will mark brainly

Semmy [17]

0647 is the answer hope I helped you out in any way

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2 years ago

Which table represents a direct variation? Table A Table B Table C Table D

NeX [460]

Table B as only in B :
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2 years ago

Read 2 more answers