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bagirrra123 [75]
1 year ago
5

Write (9m)^4 without exponents.

Mathematics
2 answers:
Elodia [21]1 year ago
7 0

(9m)^4 means 9m is multiplied by itself 4 times:

It can be written as: (9m)(9m)(9m)(9m)

Heart if it was helpful

TEA [102]1 year ago
6 0
9m*9m*9m*9m because your multiplying 9 by m four times
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Zachary’s final project for college course took a semester to write and had 95,234 words.zachary wrote 35,295 words the first mo
kap26 [50]

Answer:

40,999

Step-by-step explanation:

95,234-35,295= 59,939 and then 59,939-19,240= 40,999

5 0
2 years ago
In a trapezoid the lengths of bases are 11 and 18. The lengths of legs are 3 and 7. The extensions of the legs meet at some poin
FrozenT [24]

Answer: The length of segments between this point and the vertices of greater base are 7\frac{5}{7} and 18.

Step-by-step explanation:

Let ABCD is the trapezoid, ( shown in below diagram)

In which AB is the greater base and AB = 18 DC= 11, AD= 3 and BC = 7

Let P is the point where The extended legs meet,

So, according to the question, we have to find out : AP and BP

In Δ APB and Δ DPC,

∠ DPC ≅ ∠APB ( reflexive)

∠ PDC ≅ ∠ PAB    ( By alternative interior angle theorem)

And, ∠ PCD ≅ ∠ PBA  ( By alternative interior angle theorem)

Therefore, By AAA similarity postulate,

\triangle APB\sim \triangle D PC

Let, DP =x

⇒ \frac{3+x}{18} = \frac{x}{11}

⇒  33 +11x = 18x

⇒ x = 33/7= 4\frac{5}{7}

Thus, PD= 4\frac{5}{7}

But, AP= PD + DA

AP= 4\frac{5}{7}+3 =7\frac{5}{7}

Now, let PC =y,

⇒ \frac{7+y}{18} = \frac{y}{11}

⇒ 77 + 11y = 18y

⇒ y = 77/7 = 11

Thus, PC= 11

But, PB= PC + CB

PB= 11+7 = 18



7 0
1 year ago
Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
pychu [463]

Answer:

A=152

K= -Ln(0.5)/14

Step-by-step explanation:

You can obtain two equations with the given information:

T(14 minutes) = 114◦C

T(28 minutes)=152◦C

Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:

114=190-Ae^{-14k} (I) \\152=190-Ae^{-28k}(II)

Applying the following property of exponentials numbers in (II):

e^{a}.e^{b}=e^{a+b}

Therefore e^{-28k} can be written as e^{-14k}.e^{-14k}

152=190-Ae^{-14k}.e^{14k}

Replacing (I) in the previous equation:

152=190-76e^{-14k}

Solving for k:

Subtracting 190 both sides, dividing by -76:

0.5=e^{-14k}

Applying the base e logarithm both sides:

Ln(0.5)= -14k

Dividing by -14:

k= -Ln(0.5)/14

Replacing k in (I) and solving for A:

Ae^{-14(-Ln(0.5)/14)}=76\\Ae^{Ln(0.5)} =76\\A(0.5)=76

Dividing by 0.5

A=152

7 0
2 years ago
Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th
Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

(a)

The bid should be greater than $10,000 to get accepted by the seller. Let bid x be a continuous random variable that is uniformly distributed between

$10,000 and $15,000

The interval of the accepted bidding is [ {\rm{\$ 10,000 , \$ 15,000}], where b = $15000 and a = $10000.

The interval of the provided bidding is [$10,000,$12,000]. The probability is calculated as,

\begin{array}{c}\\P\left( {X{\rm{ < 12,000}}} \right){\rm{ = }}1 - P\left( {X > 12000} \right)\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{12000}^{15000}\\\end{array}

=1- \frac{[15000-12000]}{5000}\\\\=1-0.6\\\\=0.4

(b)  The interval of the accepted bidding is [$10,000,$15,000], where b = $15,000 and a =$10,000. The interval of the given bidding is [$10,000,$14,000].

\begin{array}{c}\\P\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{14000}^{15000}\\\end{array} P(X14000)

=1- \frac{[15000-14000]}{5000}\\\\=1-0.2\\\\=0.8

(c)

The amount that the customer bid to maximize the probability that the customer is getting the property is calculated as,  

The interval of the accepted bidding is [$10,000,$15,000],

where b = $15,000 and a = $10,000. The interval of the given bidding is [$10,000,$15,000].

\begin{array}{c}\\f\left( {X = {\rm{15,000}}} \right){\rm{ = }}\frac{{{\rm{15000}} - {\rm{10000}}}}{{{\rm{15000}} - {\rm{10000}}}}\\\\{\rm{ = }}\frac{{{\rm{5000}}}}{{{\rm{5000}}}}\\\\{\rm{ = 1}}\\\end{array}

(d)  The amount that the customer bid to maximize the probability that the customer is getting the property is $15,000, set by the seller. Another customer is willing to buy the property at $16,000.The bidding less than $16,000 getting considered as the minimum amount to get the property is $10,000.

The bidding amount less than $16,000 considered by the customers as the minimum amount to get the property is $10,000, and greater than $16,000 will depend on how useful the property is for the customer.

5 0
1 year ago
X2 + y2 = 1 x = 1 Hina's work: substitute: (1)2 + y2 = 1 simplify: 12 + y2 = 1 isolate the variable: y2 = 0 solve for y: y = 0 C
yulyashka [42]

Answer:

this is not a system of equations. there is one equation. (1,0)

Step-by-step explanation:

x^2+y^2=1

x(1)+y^2=1

1+y^2=1

y^2=0

y=0

(1,0)

3 0
1 year ago
Read 2 more answers
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