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3 years ago

Find the diagonal length of this figure

Mathematics

2 answers:

sasho [114]3 years ago

6 0

Answer:

$5\sqrt{5} \:mm$

11.18 mm

Step-by-step explanation:

length of diagonal

$=\sqrt{l^{2} +w^{2}+h^{2} } \\=\sqrt{8^{2} +6^{2}+5^{2} } \\=\sqrt{64 +36+25 } \\=\sqrt{125 } \\=5\sqrt{5} \: mm\\=11.18 mm$

LekaFEV [45]3 years ago

3 0

Answer:

the answer is,

11.18mm

by pythagoras theorem, you'll get base diagonal as 10mm, therefore the diagonal*marked red is 11.18mm.

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4 years ago

Pls answer asap 7 - 3 r = r - 4 ( 2 + r)

Marianna [84]

Answer:

no real answer

Step-by-step explanation:

distribute the -4 and combine all of the like terms on the left side = r-8-4r, then -8-3r

then we have 7-3r=-8-3r

from here, we can already tell that there's no real answer. this is because the two -3r will cancel, leaving no variable.

since 7 doesn't equal -8, there is no answer.

if, for example, the value on both sides of the equal sign were the same after the variable was eliminated, then your answer would be all real numbers

8 0

3 years ago

(3x + 5y = 7<br> { 4x - y = 5

dalvyx [7]

Answer:

Step-by-step explanation:

Solution by substitution method

3x+5y=7

and 4x-y=5

Suppose,

3x+5y=7→(1)

and 4x-y=5→(2)

Taking equation (2), we have

4x-y=5

⇒y=4x-5→(3)

Putting y=4x-5 in equation (1), we get

3x+5y=7

⇒3x+5(4x-5)=7

⇒3x+20x-25=7

⇒23x-25=7

⇒23x=7+25

⇒23x=32

⇒x=32/23

→(4)

Now, Putting x=32/23

in equation (3), we get

y=4x-5

⇒y=4(32/23)-5

⇒y=(128-115)/23

⇒y=13/23

∴y=13/23 and x=32/23

5 0

2 years ago

Which is log x+ 6 log(x+7) written as a single logarithm.

Elina [12.6K]

Answer: option d.

Step-by-step explanation:

To solve this problem you must keep on mind the properties of logarithms:

$log(b)-log(a)=log(\frac{b}{a})\\\\log(b)+log(a)=log(ba)\\\\a*log(b)=log(b)^a$

Therefore, knowing the properties, you can write the expression gven in the problem as shown below:

$5logx+6log(x+7)=logx^5+log(x+7)^6=logx^5(x+7)^6$

Then, the answer is the option d.

4 0

3 years ago

Read 2 more answers

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago