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lozanna [386]
3 years ago
9

4(2+3)-[7-[4-(8+5)]can you help me with this question

Mathematics
2 answers:
Gnesinka [82]3 years ago
7 0
4(2 + 3) - [7 - [4 - (8 + 5)]]

First, do the parentheses:

4(5) - [7 - [4 - 12]]

Then complete the rest from the inside out:

20 - [7 - (-8)]

20 - 15 = 5
natali 33 [55]3 years ago
6 0
If you simplify or evaluate the problem you get 4. You should check out math way.
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Hey guys<br>im new here<br>please solve this for me with steps!<br>ill mark as the best answer​
Vinil7 [7]

Answer:

The factors of  2(x+y)^2-9(x+y)-5 is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=>2(x+y)^2-9(x+y)-5

To Find:

The factors of the polynomial =?

Solution:

Lets assume  k = (x+y)

Then 2(x+y)^2-9(x+y)-5 can be written as 2k^2-9k-5

Now by using quadratic formula

k =\frac{-b\pm\sqrt{(b^2-4ac}}{2a}

where

a= 2

b= -9

c= -5

Substituting the values, we get

k =\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}

k =\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}

k =\frac{-(-9) \pm \sqrt{(81+40)}}{4}

k =\frac{-(-9) \pm \sqrt{(121)}}{4}

k =\frac{-(-9) \pm 11}}{4}

k= \frac{ 9 \pm 11}{4}

k =  \frac{20}{4}                         k =  \frac{-2}{4}    

k_1 =5                                      k_2 = -\frac{1}{2}

2k^2-9k-5= 2(k-5)(k+\frac{1}{2})

Solving the RHS we get

\frac{2}{2}(k-5)(2k+1)

(k-5)(2k+1)

Substituting k = x+y

((x+y)-5)(2(x+y+1)

((x+y)-5)(2x+2y+1)

5 0
3 years ago
Seven and fifty-one thousands in numbers
mihalych1998 [28]
51,007.

If you want it is thousandths,

7.051.

Bye!





8 0
3 years ago
Read 2 more answers
Cos2a= 2cos^2a-1 for all values of a<br><br> true or false?
Law Incorporation [45]
True
<span>Cos(A+B)=CosACosB-SinASinB
therefore Cos(A+A)= CosACosA - SinASinA
= Cos^2A - Sin^2A</span>
5 0
4 years ago
Read 2 more answers
Find the missing angle measure.
ycow [4]

Step-by-step explanation:

The sum of all inner angles in the shape should be 540°

(180° for triangles, 360° for squares and other simple 4-corner-shapes, the pattern is the number of corners minus 2 multiplied by 180°)

we can calculate

540-106-94-135=205

so we got 205 degrees for the two unclear corners and one of them has to be 5° greater.

x is 100°

x is 100°x+5 is 105°

(note that in the subtraction part we could have subtracted 5 more and would be left with 2x=200)

5 0
3 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
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