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2 years ago

1+4=5, 2+5=12, 3+6=21,8+11=

Mathematics

1 answer:

klemol [59]2 years ago

8 0

The answer is 40 because once you have 21 your just adding 19 and then that makes it 40.

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Triangle BCD is isosceles and BC ≅ BD. Circle A is shown. Isosceles triangle C B D has points on the circle. The lengths of C B

Masja [62]

Answer:

130

Step-by-step explanation:

since bc and bd equal the same thing, plug in every answer option until you get to 360.

(130 x 2) = 260

plus cd = 100.

260 + 100 = 360.

3 0

2 years ago

Read 2 more answers

An emu that measures 60 inches in height is 70 inches less than 5 times the height of a kakapo. What is the height of a kakapo i

Lunna [17]

<span>We can solve the following equation: 5 x - 70 = 60, where x is the height of a kakapo in inches : 5 x - 70 + 70 = 60 + 70 ( we will add 70 to the both sides of an equation ), 5 x = 130, x = 130 : 5, x = 26 inches. We can prove it: 26 * 5 - 70 = 130 - 70 = 60 inches ( the height of an emu in inches ). Answer: The height of a kakapo is 26 inches.Hope I helped! :) Cheers!</span>

7 0

2 years ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

2 years ago

Suppose that the IQs of universityâ€‹ A's students can be described by a normal model with mean 140 and standard deviation 8 poi

AnnZ [28]

Answer:

0.0266, 0.9997,0.7856

Step-by-step explanation:

Given that the IQs of universityâ€‹ A's students can be described by a normal model with mean 140 and standard deviation 8 points. Also suppose that IQs of students from university B can be described by a normal model with mean 120 and standard deviation 11. Let x be the score by A students and Y the score of B.

A) $P(X>135) = \\P(Z>0.625)\\=0.0266$