3 years ago

PleAse help please please

1 answer:

8 0

No he is not correct. For him to be correct the numbers would have to be the same but the negatives and positives would be different.

I don't know if this helps at all, probably not, but I tried. It made sense in my head but it might not make sense to other people. Sorry! Hope this works!

You might be interested in

What is the approximate area of the shaded region in the given square containing 2 semicircles? (Use * 3.14.) 10 cm 0 21.5 squar

lakkis [162]

Answer:

yes

Step-by-step explanation:

bcuz

4 0

3 years ago

Read 2 more answers

Help help help !!!!! Thanks

STatiana [176]

The next line will be 5x5=25

6 0

3 years ago

Read 2 more answers

This time the florist is making bouquets with daisies and carnations. She has 35 daisies and 42 carnations. She wants to put an

mash [69]

Answer:

7 flowers

Step-by-step explanation:divide 7 from each and you'll gat your answer

6 0

2 years ago

Please help !! Solve for x: -2 -2/3 -3/2 -3

Kryger [21]

Answer:

x = $\frac{3}{2}$

Step-by-step explanation:

Given 2 intersecting chords, then the product of the parts of one chord is equal to the product of the parts of the other chord, that is

18x = 9 × 3 = 27 ( divide both sides by 18 )

x = $\frac{27}{18}$ = $\frac{3}{2}$

8 0

3 years ago

una escalera de 12ft de longitud esta apoyada contra una pared, de modo que su base se encuentra a 4 ft de la pared a nivel del

Marrrta [24]

Answer:

Step-by-step explanation:

Altura = 11.312 pies

i) La escalera que descansa contra la pared formará un triángulo en ángulo recto.

ii) la escalera formará la hipotenusa del triángulo rectángulo dado por h = 12 pies.

iii) La base está dada por 4 pies.

iv) de la fórmula del triángulo en ángulo recto.

base = $altura = \sqrt{hipotenusa^{2}\hspace{0.15cm}- \hspace{0.15cm} base^{2} } = \sqrt{12^{2}\hspace{0.15cm} - \hspace{0.15cm}4^{2} } = \sqrt{144\hspace{0.15cm} - \hspace{0.15cm}16} = \sqrt{128} = \hspace{0.15cm}8\sqrt{2} = 8\times1.414 = 11.312 pies$

6 0

3 years ago