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2 years ago

50 POINTS!!!!!! Please help me!!! What is m<4+ m<5+ m<6? Please help me!!!

Mathematics

1 answer:

nadya68 [22]2 years ago

4 0

Answer:

360 degrees

Step-by-step explanation:

the exterior angles of the triangle is 360 degrees

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There are 2 different maps of California. The scale on the first map is 1cm to 20km. The distance from Fresno to San Francisco i

Oksi-84 [34.3K]

Answer:

75 km

Step-by-step explanation:

Given the following :

Scale on first map = 1cm to 20 km

Distance fromFresno to San Francisco = 15km on first map

Scale on 2nd map = 1cm to 100 km

1cm / 20km = 15km

Let of Distance from Fresno to San Francisco on 2nd map = d

1cm / 100km = d

1 : 20 = 15

1 : 100 = d

20/ 100 = 15/d

1/5 = 15/d

d = 15 * 5

d = 75 km

8 0

3 years ago

In Art Class, you designed a rectangle stained glass window. The

Ierofanga [76]

Answer:

Step-by-step explanation:

Draw a triangle and label the sides (picture below):

● the length is 2x+2

● and the width or the height is 6x

The perimeter of a rectangle is the sum of the sides

● P = 2x+2 + 6x + 2x+2 + 6x

Isolate the similar terms

● P = (2x+2x+6x+6x) + (2+2)

● P = 16x + 4

4 0

3 years ago

What is the volume of this triangular Pyramid photo attached

storchak [24]

To find the volume of a triangular pyramid, the formula is:

$V=\frac{1}{3}(Area\text{ }of\text{ }thebase)(height)$

So we start by finding the area of the base, which is in the shape of a right triangle in this case.

The area of a right triangle is equal to 1/2 x base x height.

$\begin{gathered} A=\frac{1}{2}bh \\ \\ A=\frac{1}{2}(5)(6) \\ \\ A=15\text{ square feet} \end{gathered}$

We use this area to find the volume.

$\begin{gathered} V=\frac{1}{3}Ah \\ \\ V=\frac{1}{3}(15)(6) \\ \\ V=30\text{ cubic feet} \end{gathered}$

The volume of the triangular pyramid is 30 cubic feet.

3 0

1 year ago

20 points

gavmur [86]

Answer:

Michael buys 800 hockey cards from jackson

6 0

2 years ago

Read 2 more answers

For fun question

ZanzabumX [31]

Consider the function $f(x)=x^{1/3}$ , which has derivative $f'(x)=\dfrac13x^{-2/3}$ .

The linear approximation of $f(x)$ for some value $x$ within a neighborhood of $x=c$ is given by

$f(x)\approx f'(c)(x-c)+f(c)$

Let $c=64$ . Then $(63.97)^{1/3}$ can be estimated to be

$f(63.97)\approxf'(64)(63.97-64)+f(64)$
$\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375$

Since $f'(x)>0$ for $x>0$ , it follows that $f(x)$ must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function $f(x)$ . This means the estimated value is an overestimation.

Indeed, the actual value is closer to the number 3.999374902...

4 0

3 years ago