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3 years ago

Which is equal to -(4 - 5)2 + 1?

Mathematics

2 answers:

Bess [88]3 years ago

7 0

Answer:

Step-by-step explanation:

Given that,

-(4 - 5)2 + 1

Not sure if the 2 at the back of the bracket is square or not..

I will solve both ways

First we will apply BODMAS, I.e solve the bracket first

-(4 - 5)2 + 1

-(-1)2 + 1

So note that - × - = +

Then,

(-1 × -2) + 1

2 + 1 = 3

Let assume it is squared

-(4 - 5)² + 1

-(-1)² + 1

-(-1 × -1) + 1

-(1) + 1

-1 + 1

So the answer is 0

I hope this help!!! cc sadiku

DiKsa [7]3 years ago

4 0

Answer:

It is equal to 3

Step-by-step explanation:

Follow PEMDAS to find the answer

Parenthasis first

- ( -1 ) * 2 + 1

No exponents

Multiplication ( left to right)

1 * 2 + 1

2 + 1

No division

Addition

No subtraction

At the end, you get 3

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Reduce to simplest form.

BigorU [14]

Convert 3 6/2 to an improper fraction so: 12/2
convert 2 4/5 to 14/5
multiply the numerators of the improper fractions so 12*14, which is 168.
multiple the denominators so 2*5, which is 10.
the new fraction is 168/10, so you divide 168 divided by 10, which is 16.8, or 16 4/5.

8 0

4 years ago

Solve the proportion 42/15=p/10

victus00 [196]

Consider the given proportion,

$\frac{42}{15}=\frac{p}{10}$

Cross multiplying in the above equation, we get

$42 \times 10 = p \times 15$

Dividing both sides of the above equation by 15, we get

$\frac{42 \times 10}{15} = p$

$p = 14 \times 2$

p = 28

So, the value of 'p' in the given proportion is '28'.

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3 years ago

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find

fenix001 [56]

Answer:

y2 = (6x + 7)/36 + (Dx + E)e^x

Step-by-step explanation:

The method of reduction of order is applicable for second-order differential equations.

For a known solution y1 of a 2nd order differential equation, this method assumes a second solution in the form Uy1 which satisfies the said differential equation. It then assumes a reduced order for U'' (w' = U'').

The differential equation becomes easy to solve, and all that is left are integration and substitutions.

Check attachments for the solution to this problem.