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Triss [41]
3 years ago
12

What shape is produced when slicing a right rectangular pyramid perpendicular to the base

Mathematics
1 answer:
vovikov84 [41]3 years ago
5 0

For the most part, the cross-section formed is a <em>trapezoid</em>, but if the slice passes through the apex of the pyramid, that shape is a <em>triangle</em>.

(Image source: MathCaptain.com)

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Help pls!!!
SVEN [57.7K]

Answer:

an = -6n + 5.

Step-by-step explanation:

This is an arithmetic sequence with first term a1 = -1 and common difference d = -6

an = a1 + d(n-1)

an = -1 + -6(n - 1)

an = -1 -6n + 6

an = -6n + 5.

3 0
3 years ago
Solve x42=37 x 42 = 3 7 using two different strategies. Explain each strategy.
Dafna11 [192]

Given equation : \frac{x}{42}=\frac{3}{7}.

Strategy 1: We can cross mutiply both sides remove fraction form.

On cross multiplication, we get

x * 7 = 3 * 42

7x = 126.

Dividing both sides by 7, we get

<h3>x = 18.</h3>

Strategy 2: We can find least common denominator(lcd) of both sides and multiplying both sides by that lcd to get rid denominators from both sides.

LCD of 42 and 7 is 42.

Therefore, multiplying both sides by 42, we get

42\times \frac{x}{42}=42\times\frac{3}{7}

x = 6 * 3

<h3>x = 18.</h3>
4 0
3 years ago
What is the slope of the line given by the equation y = 12x?
FrozenT [24]

Answer:

up 12 over 1

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Can I get help solving this graph please?
Daniel [21]
see the attached figure with the letters

1) find m(x) in the interval A,B
A (0,100)  B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100

2) find m(x) in the interval B,C
B(50,40)  C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20

3) find n(x) in the interval A,B
A (0,0)  B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x

4) find n(x) in the interval B,C
B(50,60)  C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30

5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then 
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x) 
</span>h'(x)=-36/25=-1.44

6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then 
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x) 
h'(x)=18/25=0.72 

for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

<span> h'(x) = 1.44 ------------ > not exist</span>

8 0
3 years ago
HELP WILL MARK BRAINLIEST!
denpristay [2]

Answer:

All the exponents in the algebraic expression must be non-negative integers in order for the algebraic expression to be a polynomial. As a general rule of thumb if an algebraic expression has a radical in it then it isn't a polynomial

so no its not.

3 0
2 years ago
Read 2 more answers
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