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vladimir1956 [14]
3 years ago
12

I need help finding the surface area of a net its base is 6cm and height 5cm and theres 4 triangles(50 points for who can help)​

Mathematics
2 answers:
Trava [24]3 years ago
6 0
The answer is 120
Because each triangle has an area of 30
So if you multiply 30x4 it would equal 120
nata0808 [166]3 years ago
4 0

Answer:

not sure

Step-by-step explanation:

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Denise and Donna want to exchange secrets from across a crowded whispering gallery. Recall that a whispering gallery is a room w
Yuri [45]

Answer:

34 feet

Step-by-step explanation:

From the question, we are told that:

If the room is 30 feet high at the center and 90 feet wide at the floor,

Entire width of the room = 90 feet = 2a

a = 90/2 = 45

Height of the room = b = 30 feet

Foci (c)² = a² - b²

c = √a² - b²

c = √45² - 30²

c = √1125

c = 33.541019662 feet

Approximately to the nearest whole number = 34 feet

Denise and Donna should each stand 34 ft from the outer wall so that they will be positioned at the foci of the ellipse

7 0
3 years ago
Robert bought $1.60 worth of stamps with 20 coins all in nickels and dimes. how many nickels and dimes did he use?
mrs_skeptik [129]
Nickels are 5 cents, and dimes are 10 cents.
(By luck..) eventually you can have 12 dimes, meaning you have 120 cents, or $1.20 ... 20-12=8.. meaning you used 12 dimes and are missing 8 more coins. which means you can have 8 nickels which is 40 cents. 120+40=160 cents or $1.60
Hope this helped.

3 0
4 years ago
A building casts a shadow 168 meters long. At the same time, a pole 5 meters high casts a shadow 20 meters long . What is the he
yawa3891 [41]

Answer:

42

Step-by-step explanation:

The height of the pole is 4 times shorter than the length of the shadow 20÷5=4

The height of the building must be 4 times shorter than its shadow length 168÷4=42

3 0
3 years ago
Pleaseee helppp with thissss asapppp
Natasha2012 [34]

As we know that the standard equation of circle is {\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}} , where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>

Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

{:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}

{:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}

{:\implies \quad \sf 9+16=r^{2}}

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>

Now , putting values in our standard equation ;

{:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}

<em>This is the required equation of </em><em>Circle</em>

Refer to the attachment as well !

8 0
2 years ago
Leah's parents are planning to give her a gift and want to wrap it first. The gift is in a rectangular box with a height of 3 in
jekas [21]
128 square inches of wrapping paper, since the width would be the Z Axis, That would make a Square 3X4= 12, 12X2= 24. 3X7= 21X4= 104, 104+24=128? That's what I figured. Can you send me a photo of the problem?
8 0
4 years ago
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