2 The table shows the temperature in Warsaw at

63+d=105 What is this answerrr

Allushta [10]

Answer:

42

Step-by-step explanation:

105 - 63 = 42

7 0

2 years ago

Read 2 more answers

Suzie made a mistake in the following problem. The mistake was made in Line ___. Only input the number of the first incorrect li

zalisa [80]

Answer:

Mistake in Line 2

Step-by-step explanation:

Line 1

7(6) ÷ 5 + 42

Line 2

7(6) ÷ 5 + 16

Line 3

42 ÷ 5 + 16

Line 4

42 ÷ 21

Line 5

2

To simplify any expression we use order of operation

Line 1 is 7(6) ÷ 5 + 42

Order of operation is PEMDAS

First we start with parenthesis

so we multiply 7(6) in line 2

7*6- 42

So Line 2 is 42 ÷ 5 + 42

Hence there is a mistake in Line 2

3 0

2 years ago

Solve the following equations for 0 ≤ theta ≤ 360º cosec(2 — 45°) = 2

lions [1.4K]

Step-by-step explanation:

here ,

$cosec \alpha = \frac{1}{ \sin( \alpha ) }$

now,

cosec(2-45°)=2

or,

1/sin(2-45°) =2

or,

1/sin2cos45°-cos2sin45=2

or,

$\frac{1}{ \sin(2) \frac{1}{2} - \cos(2) \frac{1}{2} } = 2$

or,

sorry I have that much qualifications to work on this question and hoping this much will make bit easy for you to solve it further

4 0

1 year ago

The diagonals of a rhombus measures 6cm and 8cm, what is the perimeter of this rhombus?

Kamila [148]

<h2><u>Diagram</u><u>:</u><u>-</u></h2>

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf A}\put(6,0.5){\sf B}\put(1.4,4.3){\sf D}\put(6.6,4.3){\sf C}\end{picture}$

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

The diagonals of the Rhombus= d1 & d2=6cm and 8cm

Let the Side=a

As we know that in a Rhombus

${\boxed{\sf side=\sqrt {\left ({\dfrac{d1}{2}}\right)^2+\left ({\dfrac {d2}{2}}\right)^2}}}$

Substitute the values

${:}\longmapsto$ $\sf a=\sqrt {\left ({\dfrac {6}{2}}\right)^2+\left ({\dfrac {8}{2}}\right)^2 }$

${:}\longmapsto$ $\sf a=\sqrt {(3)^2+(4)^2}$

${:}\longmapsto$ $\sf a=\sqrt {9+16}$

${:}\longmapsto$ $\sf a=\sqrt{25}$

${:}\longmapsto$ $\sf a=5cm$

<h2>_____________</h2><h3>Again </h3>

we know that in a Rhombus

$\boxed{\sf Perimeter=4a}$

Substitute the values

${:}\longmapsto$ $\sf Perimeter=4×5$

${:}\longmapsto$ $\sf Perimeter=20cm$

$\therefore$ Perimeter of the rectangle is 20cm.

5 0

2 years ago

If (x, y) = (3, 4) is a solution of the simultaneous equations.

Vinvika [58]

The first thing you want to do is plug in x and y into both equations:

a(3) + b(4) = 4

b(3) + a(4) = 8

rearrange to line up a’s and b’s

3a + 4b = 4

4a + 3b = 8

now you want to choose a or b and multiply each equation by a number to make them have the same amount of a’s or b’s.

4(3a + 4b = 4) = 12a + 16b = 16

3(4a + 3b = 8) = 12a + 9b = 24

Now we subtract the bottom equation from the top and solve for b:

12a + 16b - (12a + 9b) = 16 - 24

7b = -8

b = -8/7

Now we plug back in for b to one of the original equations:

3a + 4(-8/7) = 4

3a + (-32/7) = 4

3a - (32/7) = 4

3a = 4 + (32/7)

3a = (28/7) + (32/7)

3a = 60/7

a = (60/7)/3 = 20/7.

Finally, plug a and b in together to double check using the second equation.

4a + 3b = 8

4(20/7) + 3(-8/7) = ?

(80/7) - (24/7) = ?

56/7 = 8.

4 0

2 years ago