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Neko [114]
3 years ago
5

Please help, seriously confused

Mathematics
1 answer:
Rufina [12.5K]3 years ago
6 0
Original height = original width = x mm

1. She made an enlarged copy...
height = 2x
width = 2x

2. She cut off a rectangle...
height = 2x
width = \frac{2}{3} *2x= \frac{4}{3}x

3.She doubled the width...
height = 2x
width = 2*\frac{4}{3}x = \frac{8}{3}x

height * width = area

2x * \frac{8}{3}x=139 \ 968  \\ \\ \frac{16}{3}x^2= 139 \ 968 \\\\ x^2=139 \ 968: \frac{16}{3}=139 \ 968 * \frac{3}{16}=  8 \ 748*3 = 26 \ 244 \\  \\ x= \sqrt{26 \ 244} = 162 \ mm

Original height was 162 mm 
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Someone please help me!!!!!
Katarina [22]

1. Answer (D). By the law of sines, we have \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} in any \triangle ABC.

2. Answer (C). The law of cosines, c^2=a^2+b^2-2ab\cos C,accepts up to three sides and an angle as an input.

3. Answer (D). Although this triangle is right, we are not given enough information to uniquely determine its sides and angles - here, we need either one more side or one more angle.

4. Answer (D). Don't get tripped up by answer choice (C) - this is just a rearrangement of the statement of the law of cosines. In choice (D), the signs of a^2 and 2ab\cos C are reversed.

5. Answer (B). By the law of sines, we have \frac{5}{\sin 40^\circ}=\frac{3}{\sin\theta}. Solving gives \theta\approx \boxed{23^\circ},157^\circ. Note that this is the <em>ambiguous (SSA) case</em> of the law of sines, where the given measures could specify one triangle, two triangles, or none at all!

6. Answer (A). Since we know all three sides and none of the angles, starting with the law of sines will not help, so we begin with the law of cosines to find one angle; from there, we can use the law of sines to find the remaining angles.

6 0
3 years ago
The straight line 3y = 5x + h - 6 intersect the x-axis at 4k, where h and k are constants. explain h in terms of k
inn [45]

When a function intersects with the x-axis, it's y value must be 0. That means when the straight line intersects with the axis, it's at the point (4k,0), so plugging those numbers into our original equation yields:

0=20k+h-6 \implies\\ 6=20k+h \implies \\ 6-20k=h

3 0
3 years ago
If the input is -8, what was the output
Diano4ka-milaya [45]

Answer:

I think the output can be any numbers because you don't have the function to put -8 in, so I can't identify the output exactly is.

Step-by-step explanation:


5 0
3 years ago
3,000 is 10 times as much as
nalin [4]

3,000 is 10 times as much as

3000 = 10 times of what number

We need to find 10 times what number gives us 3000

Let the unknown number be x

3000 = 10 times x

3000 = 10x

To solve for x, we divide by 10 on both sides

\frac{3000}{10}=\frac{10x}{10}

300 = x

So, the unknown number is 300

3000 = 10 times 300

3000 is 10 times as much as 300


4 0
2 years ago
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