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3 years ago

I need help on the two bottom triangles please

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

5 0

Answer: 6. x=29yd 7. x-6.5cm

Step-by-step explanation:

6. All the sides of the square are equal (we know this because of the little lines on the sides), and the equation for finding the hypotanuse of a triangle is a^2+b^2=c^2. That's 20^2+21^2=x^2. That simplifies to 841^2=x^2, which then simplifies to x=29.

7. The bottom of the triangles are the same length, so each one is 2.5cm long. The angles are also the same, which means that the triangles are equal, so we have 2 similarities between the triangles, so they are both the same (You'll learn why later in math)

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If the label of a can of yams were cut off and laid flat, which expression would give the area, A, of the label with the measure

Damm [24]

I think you are missing information to solve this problem. Should there me an image attached to this question.

Either way, to solve for the label, the label would most likely me a rectangle when laid flat.

It's dimensions would be length x height.

Where height is most likely "h"
and
length is most likely the circumference (2πr)

Thus the possible answers are either:

B or C

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7 0

3 years ago

-4x + (-2) = 3x + (-5)

slamgirl [31]

Answer:

Let x = 1

LHS: -4(1)+(-2)=-4+(-2)=-6

RHS:3(1)+(-5)=3+(-5)=-2

Therefore -4x+(-2) is not equal to 3x+(-5)

7 0

3 years ago

Pls help me I don't understand

frutty [35]

Answer:

$d. \: \: \: 25 {x}^{6}$

Step-by-step explanation:

Given,

Side of the square

$= 5 {x}^{3}$

Therefore, area of the square = side × side

$= {5x}^{3} \times 5 {x}^{3}$

$= 5 \times 5 \times {x}^{3 + 3}$

$= 25 {x}^{6}$

8 0

3 years ago

Read 2 more answers

Help help help help help

Inessa [10]

Answer:

I believe it's 240°

Step-by-step explanation:

It's been awhile since I've done this but I believe that's the answer, here's a picture of a pug. Smile and have a good day!

5 0

3 years ago

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