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3 years ago

HELP ASAP! This which craft makes no sense at all :/

Mathematics

2 answers:

lord [1]3 years ago

4 0

The answer should be 89.98 cm squared

VashaNatasha [74]3 years ago

4 0

Your answer is 89.98
I hope that’s works

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What is an equation of the line that passes through the point (–1,5) and is parallel

enot [183]

Answer:

y = - x + 4

Step-by-step explanation:

The first step is to find the slope-intercept form, which is:

y=x-6.

Then, y=x+6

y=-x+4 should be the answer since it passes through the point (-1, 5)

8 0

3 years ago

If lines a and b are parallel, and m<8=150, what is the m<2

ss7ja [257]

Step-by-step explanation:

$\underline{\large{ \tt{G \: I \: V \: E \: N}}} :$

Line a & b are parallel
m ∠ 8 = 150 °

$\underline{ \large{ \tt{T \: O\: \: F \: I \: N \:D}}} :$

m ∠ 2

$\underline{ \large{ \tt{S\: O \:L \: U \: T \: I \: O \: N}}} :$

$\large{ \tt{m \: \angle \: 8 \: + \: m \: \angle \: 2 = 180 \degree}}$ [ Sum of co-interior angles ]

⤑ $\large{ \tt{150 \degree \: + m \: \angle \: 2}} = 180 \degree$

⤑ $\large{ \tt{m \angle \: 2 \: = \: 180 \degree - 150 \degree}}$

⤑ $\large{ \tt{m \: \angle \: 2 \: = \bold{ \tt{30 \degree}}}}$

$\red{ \boxed{ \boxed{ \tt{Our \: final \: answer : \boxed{ \underline{ \bold{ \tt{30 \degree}}}}}}}}$

Hope I helped ! ♡

Have a wonderful day / night ! ツ

# $\underbrace{ \tt{Carry \: On \: Learning}}$ !!

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8 0

3 years ago

Read 2 more answers

An exterior angle of an isosceles triangle has measure 150°. Find two possible sets of measures for the angles of the triangle.

nasty-shy [4]

<span>_Award brainliest if helped!
If the exterior angle of the bases is 150°, then the measure of the angle of each base is
30° and the measure of the vertex is 120°. If the exterior angle of the vertex is 150°, then the measure of the angle of each base is
75° and the measure of the vertex is 30°.</span>

5 0

3 years ago

How am I supposed to figure this out if csc Theta(0) has a radical number on the top?

Harman [31]

$\bf csc(\theta)=\cfrac{hypotenuse}{opposite}\implies csc(\theta)=\cfrac{\sqrt{7}}{2}\cfrac{\leftarrow hypotenuse=c}{\leftarrow opposite=b}\\\\
-----------------------------\\\\
\textit{so hmm using the pythagorean theorem, we can say that}\\\\
c^2=a^2+b^2\implies \pm \sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
b=opposite\\
a=adjacent
\end{cases}\\\\
\textit{we'll assume is the + of the root, since you're not}\\
\textit{given further info on it, that means the I quadrant}\\\\
$

$\bf -----------------------------\\\\
cos(\theta)=\cfrac{a}{c}\qquad sec(\theta)=\cfrac{c}{a}\qquad tan(\theta)=\cfrac{b}{a}\qquad cot(\theta)\cfrac{a}{b}
$

so just find the "adjacent" side, using the pythagorean theorem, and even though the square root can give you +/-, use the positive one, assuming the angle is in the 1st quadrant

keeping in mind, that the 2nd quadrant is feasible as well, since the opposite side of +2 can also be in the 2nd quadrant

but anyhow, use the I quadrant

7 0

3 years ago

Please anyone helppp

frutty [35]

Answer:

Step-by-step explanation:

Your reflecting on the y axis first then again on the x axis

8 0

3 years ago