Can anyone help me please

Mathematics

1 answer:

Vera_Pavlovna [14]3 years ago

5 0

Answer:

1. 5(3+9) = (5 x 3) + (5 x 9)

2. (7+8)6 = (6 x 7) + (6 x 8)

3. 51 x 9 + 51 x 31 = 459 + 1581 = 2040

4. 17 x 13 + 51 x 13 + 32 x 13 = 221 + 663 + 416 = 1300

These should all be right :)

I used PEMDAS and the distributive property.

You might be interested in

Length width height equals to 62, what is the maximum volume

joja [24]

Assuming that the given shape is a cuboid,

volume=l×w×h

v=s³ (since cuboids have same length, they have same side length)

62=s³
∛62=s
s=3.957 ≈ 4 units

Hope I helped :)

4 0

3 years ago

find the smaller of 2 consecutive integers,x, if the sum of the smaller and 6 times the larger is -29

AnnyKZ [126]

Answer:- The smaller integer is -5.

Explanation:-

Let the first consecutive integer be x and the second consecutive integer be x+1.

According to the given question:-

x+6(x+1)= -29

⇒x+6x+6= -29 [distributive law]

⇒x+6x= -29-6 [subtract 6 on the both sides]

⇒7x= -35 [combining like terms ]

⇒x= -5 [dividing 7 on both sides]

Thus the smaller integer =x= -5

larger integer =x+1= -5+1= -4

3 0

4 years ago

2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

zlopas [31]

<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

Slope:

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

Slope:

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$